Will give brainliest.

Will Give Brainliest.
Will Give Brainliest.

Answers

Answer 1
The answer is the second one
Answer 2

Answer: Assuming there isnt a fourth answer, the answer is the second choice.

Step-by-step explanation: Point A is located in the first quadrant, Point B is located at 3, -1/2 and Point C is reflected off the y axis, in the second choice.


Related Questions

14. In this picture, three straight lines intersect at a point. Form an equation in x and solve for x.

Answers

Answer:

6x = 180

x = 30

Step-by-step explanation:

How would I solve the question below? In what order would I solve it?
4 ⋅ 3 + 2 ⋅ 9 − 40

Answers

Step-by-step explanation:

You would multiply 4 and 3, and 2 and 9 separately, then add them, then subtract 40. Remember PEMDAS.

(4*3) + (2*9) - 40

12 + 18 - 40

-10

Hope that helps

First you’d multiply your steps so
(4 • 3) + (2 • 9) - 40. After multiplying, solve the equation left to right. Add the two products and subtract by 40. Gotta remember PEMDAS: Parentheses, Exponents, multiplication, division, addition, subtraction

The MD orders 50mg of an elixir to be given every 12 hours. Available is 125mg/5ml. How much should be administered every 12 hours?

Answers

FORMULA

D ( desired dose) x V ( vehicle- tablet or liquid)

H ( dose on hand)

D = Dose ordered

H = dose on hand or dose on container label

V = form and amount in which drug comes ( tablet, capsule, liquid)

D= 50mg H= 125mg V= 5ml

50 x 5

125

250 divided by 125 = 2ml

So, 2ml of elixir is administered every 12 hours :) brainliest pls

Answer:

2ml

Step-by-step explanation:

50mg of some potent agent has to be given every 12 hours.

there is a solution that has a concentration of that agent of 125mg/5ml

we need to administer some part of this solution, which we cannot (or should not) change in its structure.

that means the ratio of agent to overall solution stays the same, no matter how much of the solution we administer.

all we need to do is to transit the ratio of 125/5 to represent 50/x (maintaining the said ratio).

in other words, we need to find how many ml we need to administer, so that 50mg of the agent enter the body.

so,

125/5 = 50/x

125x/5 = 50

25x = 50

x = 50/25 = 2

2ml of the solution needs to be administered every 12 hours.

Draw a triangle ABC, where AB = 8 cm , BC = 6 cm and angle B=70^ and locate its circumcentre and draw the circumcircle.​

Answers

Step-by-step explanation:

ΔABC, where AB = 8 cm, BC = 6 cm, B = 70° Construction: (i) Draw the ∆ABC with the given measurements. (ii) Construct the perpendicular bisector at any two sides (AB and BC) and let them meet at S which is the circumcircle. (iii) S as centre and SA = SB = SC as radius, draw the circumcircle to pass through A, B, and C. Circum radius = 4.3cm .draw-triangle-abc-where-cm-bc-and-70-and-locate-its-circumcentre-and-draw-the-circumcircle

al of
10. A square field has four sprinklers that spray
in the areas represented by the circles below. If
the shaded portion represents area that is not
reached by the sprinklers, find the total area that
is not reached by the sprinklers.

Answers

Using the areas of the sqaure and of the circle, it is found that the total area that is not reached by the sprinklers is of 343.36 ft².

What is the area of a square?

The area of a square of side length l is given by:

A = l²

In this problem, we have that l = 40 ft, hence:

A = (40 ft)² = 1600 ft².

What is the area of a circle?

The area of a circle of radius r is given by:

[tex]A = \pi r^2[/tex]

In this problem, we have four circles of radius r = 10 ft, hence it's combined area, in square feet, is given by:

[tex]A_c = 4\pi (10)^2 = 400\pi = 1256.64 \text{ft}^2[/tex]

The area not reached by the sprinklers is the subtraction of the area of the square by the area of the circle, hence:

1600 - 1256.64 = 343.36 ft².

More can be learned about the area of a rectangle at https://brainly.com/question/10489198

Solve the system, or show that it has no solution. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, enter the general solution in terms of x, where x is any real number.)
20x − 80y = 100
−14x + 56y = −70
(x, y) =

Answers

Answer:

The system has an infinite set of solutions [tex](x,y) = (x, \frac{x-5}{4})[/tex]

Step-by-step explanation:

From the first equation:

[tex]20x - 80y = 100[/tex]

[tex]20x = 100 + 80y[/tex]

[tex]x = \frac{100 + 80y}{20}[/tex]

[tex]x = 5 + 4y[/tex]

Replacing on the second equation:

[tex]-14x + 56y = -70[/tex]

[tex]-14(5 + 4y) + 56y = -70[/tex]

[tex]-70 - 56y + 56y = -70[/tex]

[tex]0 = 0[/tex]

This means that the system has an infinite number of solutions, considering:

[tex]x = 5 + 4y[/tex]

[tex]4y = x - 5[/tex]

[tex]y = \frac{x - 5}{4}[/tex]

The system has an infinite set of solutions [tex](x,y) = (x, \frac{x-5}{4})[/tex]

3. The size of a red blood cell is 0.000007 m and the size of a plant
cell is 0.0000127 m. Compare these two.​

Answers

Given:

Size of a red blood cell = 0.000007 m

Size of a plant cell = 0.0000127 m

To find:

The comparison of these two values.

Solution:

We have,

Size of a red blood cell = 0.000007 m

Size of a plant cell = 0.0000127 m

Clearly, [tex]0.0000127>0.000007[/tex]. Now, the difference between these two values is:

[tex]0.0000127-0.000007=0.0000057[/tex]

Therefore, the size of a plant cell is 0.0000057 m more than the size of a red blood cell.

NEED HELP
The average amount of money spent for lunch per person in the college cafeteria is $6.75 and the standard deviation is $2.28. Suppose that 18 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.


C. For a single randomly selected lunch patron, find the probability that this

patron's lunch cost is between $7.0039 and $7.8026.

D. For the group of 18 patrons, find the probability that the average lunch cost is between $7.0039 and $7.8026.

Answers

Answer:

C.[tex]P(7.0039<x<7.8026)=0.1334[/tex]

D.[tex]P(7.0039<\bar{x}<7.8026)\approx 0.2942[/tex]

Step-by-step explanation:

We are given that

n=18

Mean, [tex]\mu=6.75[/tex]

Standard deviation, [tex]\sigma=2.28[/tex]

c.

[tex]P(7.0039<x<7.8026)=P(\frac{7.0039-6.75}{2.28}<\frac{x-\mu}{\sigma}<\frac{7.8026-6.75}{2.28})[/tex]

[tex]P(7.0039<x<7.8026)=P(0.11<Z<0.46)[/tex]

[tex]P(a<z<b)=P(z<b)-P(z<a)[/tex]

Using the formula

[tex]P(7.0039<x<7.8026)=P(Z<0.46)-P(Z<0.11)[/tex]

[tex]P(7.0039<x<7.8026)=0.67724-0.54380[/tex]

[tex]P(7.0039<x<7.8026)=0.1334[/tex]

D.[tex]P(7.0039<\bar{x}<7.8026)=P(\frac{7.0039-6.75}{2.28/\sqrt{18}}<\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}})<\frac{7.8026-6.75}{2.28/\sqrt{18}})[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=P(0.47<Z<1.96)[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=P(Z<1.96)-P(Z<0.47)[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=0.97500-0.68082[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=0.29418\approx 0.2942[/tex]

A merchant keeps marble in a cylindrical plastic container that has a diameter of 28cm and height of 35cm. A marble has a diameter of 25mm. Determine the number of marbles that can be stored in such a container if air space accounts for 20% between marbles.​

Answers

Answer:

2107 marbles can be stored in the container.

Step-by-step explanation:

Since a merchant keeps marble in a cylindrical plastic container that has a diameter of 28cm and height of 35cm, and a marble has a diameter of 25mm, to determine the number of marbles that can be stored in such a container if air space accounts for 20 % between marbles, the following calculation must be performed, knowing that the volume of a cylinder is equal to height x π x radius²:

35 x 3.14 x (28/2) ² = X

109.9 x (14 x 14) = X

109.9 x 196 = X

21,540.4 = X

In turn, the volume of each 25mm diameter marble is equal to:

25mm = 2.5cm

4/3 x 3.14 x 1.25³ = X

4.18666 x 1.953125 = X

8.1770 = X

21,540.4 x 0.8 = 17,232.32

17,232.32 / 8,177 = 2,107.41

Therefore, 2107 marbles can be stored in the container.

Assume that the breaking system of a train consists of two components connected in series with both of them following Weibull distributions. For the first component the shape parameter is 2.1 and the characteristic life is 100,000 breaking events. For the second component the shape parameter is 1.8 and characteristic life of 80,000. Find the reliability of the system after 2,000 breaking events:

Answers

Answer:

0.9984

Step-by-step explanation:

we have shape parameter for the first component as 2.1

characteristics life = 100000

for this component

we have

exp(-2000/100000)².¹

= e^-0.0002705

= 0.9997

for the second component

shape parameter = 1.8

characteristic life = 80000

= exp(-2000/80000)¹.⁸

= e^-0.001307

= 0.9987

the reliability oif the system after 2000  events

= 0.9987 * 0.9997

= 0.9984

​Thirty-six percent of customers who purchased products from an​ e-commerce site had orders exceeding 110. If ​17% of customers have orders exceeding 110 and also pay with the​ e-commerce site's sponsored credit​ card, determine the probability that a customer whose order exceeds 110 will pay with the sponsored credit card.

Answers

Answer:

The right solution is "0.5".

Step-by-step explanation:

According to the question,

P(pay with the sponsored credit card | order exceeds $110)

= [tex]\frac{P(Pay \ with \ the \ sponsored \ credit\ card\ and\ order\ exceeds\ 110)}{P(order \ exceeds \ 110)}[/tex]

= [tex]\frac{P(A \ and \ B)}{P(A)}[/tex]

By putting the values, we get

= [tex]\frac{0.17}{0.34}[/tex]

= [tex]0.5[/tex]

Thus, the above is the right solution.

Classify the following polynomials. Combine any
like terms first.
x^2+3x + 2x - 2x^2
X^3+ 4x - 4x - 4x^2
X^3+2x - X^3- 2x^2+ 3

Answers

First simplify all polynomials and rewrite them in descending exponent order.

1. [tex]-x^2+2x[/tex]

2. [tex]x^3-4x^2[/tex]

3. [tex]-2x^2+2x+3[/tex]

Now observe the terms with highest exponents in each expression, in particularly focus on their exponent value,

[tex]-x^2[/tex] with value of 2

[tex]x^3[/tex] with value of 3

[tex]-2x^2[/tex] with value of 2

The value is also known as order of polynomial and it is a way to classify polynomials.

Every order creates a family of polynomials determined by the order (which is always greater than -1)

A polynomial such as (1) and (3) have an orders of 2, which is often called quadratic order and thus the polynomials (1), (3) are classified in the same family of quadratic polynomials, these are polynomials with order of 2.

Polynomial (2) however has an order of 3, which is called cubic order. This polynomial (2) is classified in the family of cubic polynomials.

There are of course many other families, in fact, infinitely many of them because you have order 0, 1, 2, 3, and so on there are precisely [tex]\aleph_0+1[/tex] read as "aleph 0 + 1" (the number of natural numbers + 1 (because 0 is not a natural number)) of polynomial families.

The first few have these fancy names, for example:

order 0 => constant polynomial

order 1 => linear polynomial

order 2 => quadratic polynomial

order 3 => cubic polynomial

order 4 => quartic polynomial

and so on.

Hope this helps!

The movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer
Which sentence correctly compares the two numbers 5.395 and 5.385?
05.395 < 5.385
05.385 > 5.395
o 5.395 = 5.385
5.385 < 5.395
h
Submit
Pass
Don't know answer

Answers

Answer:

5.385 < 5.395

Step-by-step explanation:

Compare digits one by one starting form the left.

5.395 and 5.385

The 5s in the ones place are equal.

5.395 and 5.385

The 3s in the tenths place are equal.

5.395 and 5.385

The 9 in the hundredths place is greater than the 8 in the hundredths place, so the number with the 9 is grater than the number with the 8.

That makes the number with the 8 less than the number with the 9.

Answer: 5.385 < 5.395

What are the solutions of the quadratic equation 49x2 = 9?

A. x = 1/9 and x = -1/9
B. x = 3/7 and x = -3/7
C. x = 3/4 and x = -3/4
D. x = 9/49 and x = -9/49

Brainliest if you explain how. got stumped on this one

Answers

Answer:

B

Step-by-step explanation:

49x^2=9

solve for x

x^2= 9/49

x=± [tex]\sqrt{9/49\\}[/tex]

which is x = ±3/7 (B)

Answer: b x=1/9 and x=-1/9

Step-by-step explanation:

14. The Elizabeth Tower is 320 feet tall. At what time or times during your ride on the London Eye are you at the same height as the top of the tower? Show your work. (4 points: 2 points for finding the correct time(s), 2 points for work shown)

Answers

Answer:

Ok so on a clock there is 12 numbers where 12 is on top so at 12 am and 12 pm noon and midnight you will be at the top of the clock

Hope This Helps!!!

During the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.

To determine the time(s) during the ride on the London Eye when you are at the same height as the top of the Elizabeth Tower (commonly known as Big Ben), we need to consider the height of the London Eye and its rotational motion.

Given that the Elizabeth Tower is 320 feet tall, we need to find the position of the London Eye when its height aligns with the top of the tower.

The London Eye has a height of 443 feet, and it completes one full rotation in approximately 30 minutes (or 1800 seconds). This means that it moves at a constant angular velocity of 360 degrees per 1800 seconds.

To find the time(s) when the heights align, we can set up a proportion:

(Height of the Elizabeth Tower) / (Height of the London Eye) = (Angle covered by the London Eye) / 360 degrees

Substituting the given values:

320 / 443 = (Time to align) / 1800

Simplifying the equation:

(Time to align) = (320 / 443) * 1800

Calculating the value:

(Time to align) ≈ 1303.16 seconds

Converting the time to minutes and seconds:

(Time to align) ≈ 21 minutes and 43.16 seconds

Therefore, during the ride on the London Eye, you will be at the same height as the top of the Elizabeth Tower at approximately 21 minutes and 43.16 seconds after the start of the ride.

To know more about  London Eye. here

https://brainly.com/question/16401602

#SPJ2

HELP AGAIN
235 ≤-8(1+5x)+3
i need the steps as well

Answers

Answer:

x ≤ -6

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right  

Equality Properties

Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of Equality

Step-by-step explanation:

Step 1: Define

Identify

235 ≤ -8(1 + 5x) + 3

Step 2: Solve for x

[Subtraction Property of Equality] Subtract 3 on both sides:                        232 ≤ -8(1 + 5x)[Division Property of Equality] Divide -8 on both sides:                                -29 ≥ 1 + 5x[Subtraction Property of Equality] Subtract 1 on both sides:                         -30 ≥ 5x[Division Property of Equality] Divide 5 on both sides:                                  -6 ≥ xRewrite:                                                                                                             x ≤ -6

Step-by-step explanation:

To solve for x, make sure you move everything else to the other side of the ≤ sign.

So,

[tex]235\leq -8(1+5x)+3\\232\leq -8-40x\\240\leq -40x\\-6\geq x[/tex]

* Remember that the sign changes anytime you divide by a negative number!

So your answer is:

[tex]x\leq -6[/tex], x is less than or equal to -6.

Find the missing term in the following pattern.


1984, 992, 496, blank space, 124, 62

Answers

Answer:

248

Step-by-step explanation:

common ratio for two consecutive terms is 2/1

for eg: 1984÷992 =2

992÷ 496 = 2

124÷ 62 = 2

that means 124 ×2 = 248 Answer

the answer is 248. good luck!


Solve the equation for x.
2/3x-1/9x+5=20

Answers

Answer:

x = 27

Step-by-step explanation:

I'm assuming the equation looks like this:

[tex]\frac{2}{3}x-\frac{1}{9}x+5=20[/tex]

Here's how to solve for x:

[tex]\frac{2}{3}x-\frac{1}{9}x+5=20[/tex]

(subtract 5 from both sides)

[tex]\frac{2}{3}x-\frac{1}{9}x=15[/tex]

(Find the GCF of 3 and 9, which is 3. Multiply 2/3 by 3/3. You get 6/9)

[tex]\frac{6}{9}x-\frac{1}{9}x=15[/tex]

(add like terms)

[tex]\frac{5}{9}x=15[/tex]

(multiply 9/5 to both sides, which is the same as dividing both sides by 5/9)

x = 27

Hope it helps (●'◡'●)

Help ask anyone have any more answers for the eye level program

Answers

Answer:

1) -[tex]\sqrt{32}[/tex]

2) -[tex]\sqrt{108}[/tex]

3) -[tex]\sqrt{80}[/tex]

4) -[tex]\sqrt{112}[/tex]

5) -[tex]\sqrt{40}[/tex]

6) -[tex]\sqrt{99}[/tex]

7) -[tex]\sqrt{50}[/tex]

8) -[tex]\sqrt{150}[/tex]

Step-by-step explanation:

-4√2= -√32-6√3= -√36-4√5= -√80-4√7= -√112-2√10= -√40-3√11= -√99-5√2= -√50-5√6= -√150

please mark this answer as brainlist

What is net cash flow

Answers

Net cash flow is a profitability metric that represents the amount of money produced or lost by a business during a given period.

Can you please help me with this question

Answers

Hirap nyan ah hahahahah

Police Chase: A speeder traveling 40 miles per hour (in a 25 mph zone) passes a stopped police car which immediately takes off after the speeder. If the police car speeds up steadily to 55 miles/hour in 10 seconds and then travels at a steady 55 miles/hour, how long and how far before the police car catches the speeder who continued traveling at 40 miles/hour

Answers

Answer:

a. 18.34 s b. 327.92 m

Step-by-step explanation:

a. How long before the police car catches the speeder who continued traveling at 40 miles/hour

The acceleration of the car a in 10 s from 0 to 55 mi/h is a = (v - u)/t where u = initial velocity = 0 m/s, v = final velocity = 55 mi/h = 55 × 1609 m/3600 s = 24.58 m/s and t = time = 10 s.

So, a =  (v - u)/t =  (24.58 m/s - 0 m/s)/10 s = 24.58 m/s ÷ 10 s = 2.458 m/s².

The distance moved by the police car in 10 s is gotten from

s = ut + 1/2at² where u = initial velocity of police car = 0 m/s, a = acceleration = 2.458 m/s² and t = time = 10 s.

s = 0 m/s × 10 s + 1/2 × 2.458 m/s² (10)²

s = 0 m + 1/2 × 2.458 m/s² × 100 s²

s = 122.9 m

The distance moved when the police car is driving at 55 mi/h is s' = 24.58 t where t = driving time after attaining 55 mi/h

The total distance moved by the police car is thus S = s + s' = 122.9 + 24.58t

The total distance moved by the speeder is S' = 40t' mi = (40 × 1609 m/3600 s)t' =  17.88t' m where t' = time taken for police to catch up with speeder.

Since both distances are the same,

S' = S

17.88t' = 122.9 + 24.58t

Also, the time  taken for the police car to catch up with the speeder, t' = time taken for car to accelerate to 55 mi/h + rest of time taken for police car to catch up with speed, t

t' = 10 + t

So, substituting t' into the equation, we have

17.88t' = 122.9 + 24.58t

17.88(10 + t) = 122.9 + 24.58t

178.8 + 17.88t = 122.9 + 24.58t

17.88t - 24.58t = 122.9 - 178.8

-6.7t = -55.9

t = -55.9/-6.7

t = 8.34 s

So, t' = 10 + t

t' = 10 + 8.34

t' = 18.34 s

So, it will take 18.34 s before the police car catches the speeder who continued traveling at 40 miles/hour

b. how far before the police car catches the speeder who continued traveling at 40 miles/hour

Since the distance moved by the police car also equals the distance moved by the speeder, how far the police car will move before he catches the speeder is given by S' = 17.88t' = 17.88 × 18.34 s = 327.92 m

Find dy/dx of the function y = √x sec*-1 (√x)​

Answers

Hi there!

[tex]\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}[/tex]

[tex]y = \sqrt{x} * sec^{-1}(-\sqrt{x}})[/tex]

Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

Thus:

f(x) = √x

g(x) = sec⁻¹ (√x)

[tex]\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}[/tex]

Simplify:

[tex]\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}[/tex]

[tex]\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}[/tex]

Answer:

[tex]\displaystyle y' = \frac{arcsec(\sqrt{x})}{2\sqrt{x}} + \frac{1}{2|\sqrt{x}|\sqrt{x - 1}}[/tex]

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]:                                                                           [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]Exponential Rule [Root Rewrite]:                                                                 [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:                                                                             [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]

Derivative Rule [Chain Rule]:                                                                                 [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Arctrig Derivative:                                                                                                 [tex]\displaystyle \frac{d}{dx}[arcsec(u)] = \frac{u'}{|u|\sqrt{u^2 - 1}}[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle y = \sqrt{x}sec^{-1}(\sqrt{x})[/tex]

Step 2: Differentiate

Rewrite:                                                                                                         [tex]\displaystyle y = \sqrt{x}arcsec(\sqrt{x})[/tex]Product Rule:                                                                                                [tex]\displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})][/tex]Chain Rule:                                                                                                     [tex]\displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[\sqrt{x}] \bigg][/tex]Rewrite [Exponential Rule - Root Rewrite]:                                                 [tex]\displaystyle y' = \frac{d}{dx}[x^\bigg{\frac{1}{2}}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[x^\bigg{\frac{1}{2}}] \bigg][/tex]Basic Power Rule:                                                                                         [tex]\displaystyle y' = \frac{1}{2}x^\bigg{\frac{1}{2} - 1}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{1}{2} - 1} \bigg][/tex]Simplify:                                                                                                         [tex]\displaystyle y' = \frac{1}{2}x^\bigg{\frac{-1}{2}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{-1}{2}} \bigg][/tex]Rewrite [Exponential Rule - Rewrite]:                                                           [tex]\displaystyle y' = \frac{1}{2x^\bigg{\frac{1}{2}}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2x^\bigg{\frac{1}{2}}} \bigg][/tex]Rewrite [Exponential Rule - Root Rewrite]:                                                 [tex]\displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2\sqrt{x}} \bigg][/tex]Arctrig Derivative:                                                                                         [tex]\displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{1}{|\sqrt{x}|\sqrt{(\sqrt{x})^2 - 1}} \cdot \frac{1}{2\sqrt{x}} \bigg][/tex]Simplify:                                                                                                         [tex]\displaystyle y' = \frac{arcsec(\sqrt{x})}{2\sqrt{x}} + \frac{1}{2|\sqrt{x}|\sqrt{x - 1}}[/tex]

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

-9(m + 2) + 406 - 7m)

Answers

Answer:

-9(m+2)+406-7m)

=-16+388

Solve algebraically.
6(t-2) + 15t < 5(5 + 3t)

With work shown please!!​

Answers

Step-by-step explanation:

6t-12+15t | 25+15t

21t-12 | 25+15t

21t-12 < 25+15t

hence proved..

Answer:

21t - 12 < 25 + 15t

Step-by-step explanation:

6( t - 2 ) + 15t < 5 ( 5 + 3t )

Distribute .

6t - 12 + 15t < 25 + 15t

Combine like terms.

21t - 12 < 25 + 15t.

Hence , Proved.

95, 86, 78, 71, 65, 60 _____

Answers

Answer:

hello there here is your answer

51 is your next term.

Step-by-step explanation:

you are subtracting 9 from each number

95-9= 86

86-9=78

78-9=65

65-9=60

60-9=51

so on and so on

Hope this help

have a good day

bye

Step-by-step explanation:

[tex]here \: is \: your \: solution: - \\ \\ given \: number \: = 95.86.78.71.65.60 \\ \\ = > 95 - 9 = 86 \\ \\ = > 86 - 8 = 78 \\ \\ = > 78 - 7 = 71 \\ \\ = > 71 - 6 = 65 \\ \\ = > 65 - 5 = 60 \\ \\ \: now \: follow \: the \: sequence \: \\ \\ subtract \: 4 \: from \: 60 \\ \\ = > 60 - 4 = 56 \\ \\ = > \: \: 56 \: \:( ANSWER✓✓✓)[/tex]

Can the range of a function be written like this {6,7,8,10} instead of like this [tex]6\leq x\leq 10[/tex]?

Answers

Answer:

No unless x is being used to define only elements of an integer set.

Step-by-step explanation:

No, not in general unless x is defined as a integer or a subset of the integers like the naturals, whole numbers....

Usually 6<=x<=10 means all real numbers between 6 and 10, inclusive. This means example that 6.6 or 2pi are in this set with infinitely other numbers that I can't name.

{6,7,8,9,10} just means the set containing the numbers 6,7,8,9,10 and that's only those 5 numbers.

What is the correct line graph for y=3x+5?

Answers

Answer:

The equation y=−3x+5 is in slope intercept form, and represents a straight line in which -3 is the slope, and 5 is the y -intercept.

I’m stuck please help .

Answers

Answer:

me too

Step-by-step explanation:

me too


Solve (x - 5)2 = 3.

Answers

Answer:

x = 5±√3

Step-by-step explanation:

Equation: (x-5)² = 3

Step 1: Take the square root of both side of the equation

√(x-5)² = ±√3

x-5 = ±√3

Step 2: add 5 to both side of the equation

x-5+5 = 5±√3

x = 5±√3

Hence, from the options above, the right answer is

B. x = 5±√3

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