Answer:
Hello,
P=(30,6)
Step-by-step explanation:
[tex]x=6t^2+6\\y=t^3-2\\\\\dfrac{dx}{dt}= 12t\\\dfrac{dy}{dt}= 3t^2\\\\\dfrac{dy}{dx} =\dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } =\dfrac{3t^2}{12t} =\dfrac{t}{4} \\\\\dfrac{t}{4} =\dfrac{1}{2} \Longrightarrow t=2\\\\\\x=6t^2+6=6*2^2+6=30\\\\y=t^3-2=2^2-2=8-2=6\\\\\\Tangence\ point=(30,6)\\[/tex]
The point on the curve x = 6t² + 6, y = t³ - 2 where the tangent line have slope 1/2 is (30, 6).
How to depict the point on the curve?From the information given, x = 6t² + 6, y = t³ - 2. We'll find the first order derivative of x and y which will be:
dx/dt = 12t
dy/dt = 3t²
Therefore, 3t²/12t = t/4, t = 2.
We'll put the value of t back into the equations.
x = 6t² + 6,
x = 6(2)² + 6
x = 24 + 6 = 30
y = t³ - 2.
y = (2)³ - 2
y = 8 - 2 = 6
In conclusion, the correct options is (30, 6).
Learn more about slope on:
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Which is a perfect square?
6’1
6’2
6’3
6’5
Answer:
6'2
Step-by-step explanation:
Question 8 plz show ALL STEPS
Answer:
Substitute the functions and the value of the functions.
Step-by-step explanation:
Doing all will be long, so i'll present a and d
Here,(no a)
f(x)=3x-1, g(x)=x^2+2
Now,
f(g(x))=f(x^2+2)=3(x^2+2)-1=3x^2+6-1=3x^2+5
g(f(x))=g(3x-1)=(3x-1)^2+2=9x^2-6x+1+2=9x^2-6x+3
Here, (no d)
f(x)=x^2-9, g(x)=√(x+4)
Now,
f(g(x))=f(√(x+4))=(√(x+4))^2-9=x+4-9=x-5
g(f(x))=g(x^2-9)=√(x^2-9+4)=√(x^2-5)
Show all work to identify the asymptotes and zero of the function f(x)=6x/x^2-36
9514 1404 393
Answer:
asymptotes: x = ±6
zero: x = 0
Step-by-step explanation:
The vertical asymptotes of the function will be at the values of x where the denominator is zero. The denominator is x^2 -36, so has zeros for values of x that satisfy ...
x^2 -36 = 0
x^2 = 36
x = ±√36 = ±6
The vertical asymptotes of the function are x = -6 and x = +6.
__
The zero of the function is at the value of x that makes the numerator zero. This will be the value of x that satisfies ...
6x = 0
x = 0 . . . . . divide by 6
The zero of the function is x=0.
__
As a check on this work, we have had a graphing calculator graph the function and identify the zero.
The 4th of an AP is 15 and the 9th term is 35. find the 15th term
Consecutive terms in this sequence are separated by a constant c, so if the 4th term is 15, then the next terms would be
5th: 15 + c
6th: (15 + c) + c = 15 + 2c
7th: (15 + 2c) + c = 15 + 3c
and so on. More generally, since any given number in the sequence depends on the number that came before it, we can write the n-th term in terms of the 4th term,
n-th: 15 + (n - 4) c
Then the 9th term in the sequence is
15 + (9 - 4) c = 35
and solving for c gives
15 + 5c = 35 ==> 5c = 20 ==> c = 4
Then the 15th term would be
15 + (15 - 4)×4 = 15 + 11×4 = 15 + 44 = 59
HELP ASAP PLEASE I WILL MARK BRAINLEST
Show all work to identify the asymptotes and zero of the function f of x equals 6 x over quantity x squared minus 36.
Answer:
vertical asymptotes
x=6, x=-6
horizontal asymptotes
y=0
zeros (0,0)
Step-by-step explanation:
f(x) = 6x / ( x^2 - 36)
First factor
f(x) = 6x / ( x-6)(x+6)
Since nothing cancels
The vertical asymptotes are when the denominator goes to zero
x-6 = 0 x+6=0
x=6 x= -6
Since the numerator has a smaller power than the denominator (1 < 2), there is an asymptote at y = 0
To find the zeros, we find where the numerator = 0
6x=0
x=0
[tex]\\ \rm\Rrightarrow y=\dfrac{6x}{x^2-36}[/tex]
The h orizontal asymptote
As x has less degree than x²
y=0 is a asymptoteVertical asymptote
x²-36=0x²=36x=±6Manatees can swim in water up to 20 feet deep. Write an expression that represents the depth d, that a manatee can swim
Answer:
0 ≤ d ≤ 20
Step-by-step explanation:
You mention that Manatees can swim in water up to 20 feet deep. So, this means that the largest depth that he can swim is 20 feet, not more than this. Also, keep in mind that the depth can't be negative, so ----> 0 ≤ d ≤ 20 feet
We want to write an expression (an inequality actually) that defines the depth at which a manatee can swim. The inequality is: 0ft ≤ d ≤ 20ft.
We know that the manatees can swim in water up to 20 feet deep. This represents the maximum deep at which manatees can swim, the minimum is trivial, it would be 0ft (when the manatees are on the surface of the water).
Then we can write the inequality:
0ft ≤ d ≤ 20ft.
This gives the range of possible values of d, depth at which the manatee can swim.
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What is the base and height of parallelogram S?
Fill in the blanks.
(3b^3)^2 = _b^_
We can seperate (3b³) into two different parts, the constant and the variable.
The constant (3) and the variable (b) can both be squared and multiplied to get the correct answer, so:
3² = 9
(b³)² = [tex]b^{6}[/tex]
So, [tex](3b^{3})^{2} = 9b^{6}[/tex]
Dividing integers
7. (-154) ➗ (-14) =
11. (-40) ➗10=
15. 90 ➗ (-15)=
16. 108 ➗ (-9)=
17. (-48) ➗ (-6)=
18. (-105) ➗ 7=
first we shall learn the rules.when numbers with same sign are divided it gives pisitive sign but, when numbers of different signs are divided it gives negetive sign.
here,
7. (-154) ➗ (-14) =11
11. (-40) ➗10=-4
15. 90 ➗ (-15)=-6
16. 108 ➗ (-9)=-12
17. (-48) ➗ (-6)=8
18. (-105) ➗ 7=-15
hope it helps you......
Instructions: Find the measure of the indicated angle to the nearest degree.
Answer:
? = 13.6
Step-by-step explanation:
Let the unknown angle be y
so
tan y= p/b
tan y =8/33
y = tan‐¹(8/33)
y = 13.62699486
y = 13.6
create a graph of 4.95 + 3.99
Answer:
????
Step-by-step explanation:
as in y = 4.95 + 3.99 or points? if so just draw a horizontal line at 8.94
Describe what is the most difficult part of solving equations, for you personally.
What do you personaly feel like is most dificult.
For me its rembering minus signs
Cho hình hộp chữ nhật ABCD A B C D
Answer:
A B C D
A×B×C×D
3×3×3×6
162
PLEASE HELP ASAP
Solve the inequality [tex]\sqrt[3]{x+4} \ \textgreater \ \sqrt[2]{-x}[/tex]
A) x < 2
B) x > 2
C) x > –2
D) x < –2
find the area of the shaded regions. ANSWER IN PI FORM AND DO NOT I SAID DO NOT WRITE EXPLANATION
Answer: 18π
okokok gg
Step-by-step explanation:
Here angle is given in degree.We have convert it into radian.
[tex] {1}^{\circ} =( { \frac{\pi}{180} } )^{c} \\ \therefore \: {80}^{\circ} = ( \frac{80\pi}{180} ) ^{c} = {( \frac{4\pi}{9} })^{c} \: = \theta ^{c} [/tex]
radius r = 9 cmArea of green shaded regions = A
[tex] \sf \: A = \frac{1}{2} { {r}^{2} }{ { \theta}^{ c} } \\ = \frac{1}{2} \times {9}^{2} \times \frac{4\pi}{9} \\ = 18\pi \: {cm}^{2} [/tex]
Instructions: Given the following constraints, find the maximum and minimum values for
z
.
Constraints: 2−≤124+2≥0+2≤6 2x−y≤12 4x+2y≥0 x+2y≤6
Optimization Equation: =2+5
z
=
2
x
+
5
y
Maximum Value of
z
:
Minimum Value of
z
:
Answer:
z(max) = 16
z(min) = -24
Step-by-step explanation:
2x - y = 12 multiply by 2
4x - 2y = 24 (1)
4x + 2y = 0 add equations
8x = 24
x = 3
4(3) + 2y = 0
y = -6
so (3, -6) is a common point on these two lines
z = 2(3) + 5(-6) = -24
4x - 2y = 24 (1)
x + 2y = 6 add equations
5x = 30
x = 6
6 + 2y = 6
y = 0
so (6, 0) is a common point on these two lines
z = 2(6) + 5(0) = 12
4x + 2y = 0 multiply by -1
-4x - 2y = 0
x + 2y = 6 add equations
-3x = 6
x = -2
-2 + 2y = 6
y = 4
so (-2, 4) is a common point on these two lines
z = 2(-2) + 5(4) = 16
when 18 is subtracted from six times a certain number the result is 96 what is the number
Let the number be x
ATQ
[tex]\\ \sf\twoheadrightarrow 6x-18=96[/tex]
[tex]\\ \sf\twoheadrightarrow 6x=96+18[/tex]
[tex]\\ \sf\twoheadrightarrow 6x=112[/tex]
[tex]\\ \sf\twoheadrightarrow x=\dfrac{112}{6}[/tex]
[tex]\\ \sf\twoheadrightarrow x=7[/tex]
Abigail buys two cartons of strawberries. One carton has 191919 berries and the other carton has 262626 berries. She wants to divide the berries into bags so there are exactly 666 berries in each bag.
How many bags will have 666 berries?
Answer:
682
Step-by-step explanation:
191,919 + 262,626
454545 ÷ 666 = 682.5
Thus meaning 682 bags will have 666 berries and one bag will have 333 berries.
Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)
Answer:
A)
[tex]k=0[/tex]
B)
[tex]\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}[/tex]
C)
[tex]\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}[/tex]
Step-by-step explanation:
We are given the function:
[tex]\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}[/tex]
A)
Given that h(1) = 20, we want to find k.
h(1) = 20 means that h(x) = 20 when x = 1. Substitute:
[tex]\displaystyle (20) = 20e^{k(1)}[/tex]
Simplify:
[tex]1= e^k[/tex]
Anything raised to zero (except for zero) is one. Therefore:
[tex]k=0[/tex]
B)
Given that h(1) = 40, we want to find 2k + 1.
Likewise, this means that h(x) = 40 when x = 1. Substitute:
[tex]\displaystyle (40) = 20e^{k(1)}[/tex]
Simplify:
[tex]\displaystyle 2 = e^{k}[/tex]
We can take the natural log of both sides:
[tex]\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}[/tex]
By definition, ln(e) = 1. Hence:
[tex]\displaystyle k = \ln 2[/tex]
Therefore:
[tex]2k+1 = 2\ln 2+ 1 \approx 2.3863[/tex]
C)
Given that h(1) = 10, we want to find k - 3.
Again, this meas that h(x) = 10 when x = 1. Substitute:
[tex]\displaystyle (10) = 20e^{k(1)}[/tex]
Simplfy:
[tex]\displaystyle \frac{1}{2} = e^k[/tex]
Take the natural log of both sides:
[tex]\displaystyle \ln \frac{1}{2} = k\ln e[/tex]
Therefore:
[tex]\displaystyle k = \ln \frac{1}{2}[/tex]
Therefore:
[tex]\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931[/tex]
5765865876+5737555586=
Answer:
5765865876+5737555586=11503421462
One book is 4cm thick, find out how many such books can be placed in a space of 53cm.
Find the final amount of money in an account if $7, 200 is deposited at 2.5 % interest compounded
quarterly (every 3 months) and the money is left for 9 years.
The final amount is $
Round answer to 2 decimal places
The final amount is $7,615.27
A = P(1 + r/n)^t
Where,
A = Final amount
P = principal = $7, 200
r = interest rate = 2.5% = 0.025
n = number of periods = 4
t = time = 9 years
A = P(1 + r/n)^t
= 7,200(1 + 0.025/4)^9
= 7,200(1 + 0.00625)^9
= 7,200(1.00625)^9
= 7,200(1.0576769512798)
= 7,615.2740492152
Approximately,
A = $7,615.27
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A canoeist paddled down a river a distance of 2 miles in 45 minutes. Paddling up-stream on his return, it took him 90 minutes. Find the rate of the canoe in still water.
There are 768 beds in a hospital.
Each floor has 64 beds.
How many floors are there?
Answer:
12 floors
Step-by-step explanation:
768 ÷ 64 = 12.
Answer:
12
Step-by-step explanation:
768 divided by 64 =12
Find the lengths the missing side
Answer:
Short leg = x
Longer leg = 12
Hypotenuse = y
Short leg = 4√3
longer leg = 12
Hypotenuse = 8√3
Answered by GAUTHMATH
51.Tandin Dorji was married to five women. First woman had three
daughters and five sons and the youngest wife had two sons. Two
of the remaining wives had one son each. If the ratio of children of
5th wife was 1:3 with the children of other wives. How many
children does Tandin have
Answer:
Tandin has 16 children.
Step-by-step explanation:
Total of children:
3+5 = 8(first woman)
2(youngest wife)
1 + 1 = 2(two of the remaining wives)
So
8 + 2 + 2 = 12
If the ratio of children of 5th wife was 1:3 with the children of other wives.
Thus the 5th wife has 12/3 = 4 children.
How many children does Tandin have?
12 + 4 = 16
Tandin has 16 children.
Given: 3x+11=y, solve for x if y = 29
answer is 6
Step-by-step explanation:
3x+11=y
y=29
3x+11=29
3x=29-11
3x=18
x=18÷3
x=6
Answer:6
Step-by-step explanation:
3x+11=29
3x=29-11
3x=18
X=18/3
X=6
Can you please help me
Answer:
you will add the numerator and the denominator and or you look for lowest common factor
A manufacturer claims that its drug test will detect steroid use (that is, show positive for an athlete who uses steroids) 95% of the time. Further, 15% of all steroid-free individuals also test positive. 10% of the rugby team members use steroids. Your friend on the rugby team has just tested positive. The correct probability tree looks like
Answer:
The probability tree is;
0.95 [tex](+)[/tex]
[tex](S)[/tex]
0.1 0.05 [tex](-)[/tex]
[ P ]
0.9 0.15 [tex](+)[/tex]
[tex](S_{no})[/tex]
0.85 [tex](-)[/tex]
Step-by-step explanation:
Given the data in the question;
10% of the rugby team members use steroids
so Probability of using steroid; P( use steroid ) = 10% = 0.10
Probability of not using steroid; P( no steroid use ) = 1 - 0.10 = 0.90
Since the test show positive for an athlete who uses steroids, 95% of the time.
Probability of using steroids and testing positive = 95% = 0.95
Probability of using steroids and testing Negative = 1 - 0.95 = 0.05
Also from the test, 15% of all steroid-free individuals also test positive.
so
Probability of not using steroids and testing positive = 15% = 0.15
Probability of not using steroids and testing negative = 1 - 0.15 = 0.85
To set up the probability tree, Let;
[tex](S)[/tex] represent steroid use
[tex](S_{no})[/tex] represent no steroid use
[tex](+)[/tex] represent test positive
[tex](-)[/tex] represent test negative
so we have;
0.95 [tex](+)[/tex]
[tex](S)[/tex]
0.1 0.05 [tex](-)[/tex]
[ P ]
0.9 0.15 [tex](+)[/tex]
[tex](S_{no})[/tex]
0.85 [tex](-)[/tex]
Aaron Lloyd what is a?
Answer:
Rugby lawyer
Step-by-step explanation:
Aaron is a partner in the firm’s dispute resolution division. He advises clients on a range of litigious and risk related matters, with particular expertise in the areas of corporate misconduct, white collar criminal and regulatory affairs, sports law and employment law. Aaron leads our sports law practice, and is a member of the firm’s health and safety, public law, and organisational integrity teams.
Well regarded by clients for his ability to analyse and strategise complex situations, Aaron is internationally recognised for his ability to implement pragmatic and commercial strategies to minimise risk and create opportunity. This ability has resulted in clients avoiding significant litigation and commercial consequences.
Aaron is an experienced advocate, having argued cases in the District Court, High Court, Employment Court, the Court of Appeal and Supreme Court of New Zealand, along with numerous tribunals.
He is recognised by international legal directories including by Chambers & Partners (Asia Pacific), Who’s Who Legal, Expert Guides, Best Lawyers and Doyles.
Before joining MinterEllisonRuddWatts Aaron practiced as a barrister with Paul Davison QC, and has lectured at the University of Auckland.