Can I please get asap

Answer:

[tex]14[/tex]

Step-by-step explanation:

Given

[tex]\displaystyle \sum_{k=0}^3k^2[/tex]

Let's break down each part. The input at the bottom, in this case [tex]k=0[/tex], is assigning an index [tex]k[/tex] at a value of [tex]0[/tex]. This is the value we should start with when substituting into our equation.

The number at the top, in this case 3, indicates the index we should stop at, inclusive (meaning we finish substituting that index and then stop). The equation on the right, in this case [tex]k^2[/tex], is the equation we will substitute each value in. After we substitute our starting index, we'll continue substituting indexes until we reach the last index, then add up each of the outputs produced.

Since [tex]k=0[/tex] is our starting index, start by substituting this into [tex]k^2[/tex]:

[tex]0^2=0[/tex]

Now continue with [tex]k=1[/tex]:

[tex]1^1=1[/tex]

Repeat until we get to the ending index, [tex]k=3[/tex]. Remember to still use [tex]k=3[/tex] before stopping!

Substituting [tex]k=2[/tex]:

[tex]2^2=4[/tex]

Substituting [tex]k=3[/tex]:

[tex]3^2=9[/tex]

Since 3 is the index we end at, we stop here. Now we will add up each of the outputs:

[tex]0+1+4+9=\boxed{14}[/tex]

Therefore, our answer is:

[tex]\displaystyle \sum_{k=0}^3k^2=0+1+4+9=\boxed{14}[/tex]

Answer:

14

Step-by-step explanation:

∑3/k=0 k^2

Let k=0

0^2 =0

Let k = 1

1^2 =1

Let k =2

2^2 = 4

Let k = 3

3^2 = 9

0+1+4+9 = 14

9514 1404 393

Answer:

  (d)  27.4%

Step-by-step explanation:

The desired percentage is ...

  (juniors for Kato)/(total juniors) × 100%

  =  129/(129 +194 +147) × 100%

  = (129/470) × 100% ≈ 27.4%

About 27.4% of juniors voted for Kato.

9514 1404 393

Answer:

  44.66 in

Step-by-step explanation:

The side opposite the marked angle is given, and the side adjacent to it is the one wanted. The relevant trig relation is ...

  Tan = Opposite/Adjacent

Solving for the Adjacent side, we find ...

  Adjacent = Opposite/Tan

  PQ = (29 in)/tan(33°) ≈ 44.66 in

9514 1404 393

Answer:

  y = -x +3

  m = -1

  b = 3

Step-by-step explanation:

To solve the given equation for y, divide by the coefficient of y.

  (-3y)/(-3) = (3x -9)/(-3)

  y = -x +3

__

The slope is the x-coefficient, M = -1.

__

The y-intercept is the added constant, B = 3.

__

Both equations are graphed in the attachment. Texture has been added to the original so you can see the graphs are the same line.

Step-by-step explanation:

Any number that can be divided by 2 without having remainder is considered an even number.

I hope it helped U

stay safe stay happy

a. The linear equation for the first point (-4,1) is 16A-4B+C=1

b. The linear equation for the second point (-2, 0.94) is 4A-2B+C=0.94

c. The linear equation for the third point (0,1) is 0A+0B+C=1

d. The matrix equation looks like this:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

e. The inverse of the coefficient matrix looks like this:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

f. The equation of the parabola is: [tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

a. In order to build a linear equation from the given points, we need to substitute them into the general form of the equation.

Let's take the first point (-4,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](-4)^{2}A+(-4)B+C=1[/tex]

which yields:

[tex]16A-4B+C=1[/tex]

b. Let's take the second point (-2,0.94). When substituting it into the general form of the quadratic equation we end up with:

[tex](-2)^{2}A+(-2)B+C=0.94[/tex]

which yields:

[tex]4A-2B+C=0.94[/tex]

c. Let's take the third point (0,1). When substituting it into the general form of the quadratic equation we end up with:

[tex](0)^{2}A+(0)B+C=1[/tex]

which yields:

[tex]0A+0B+C=1[/tex]

d. A matrix equation consists on three matrices. The first matrix contains the coefficients (this is the numbers on the left side of the linear equations). Make sure to write them in the right order, this is, the numbers next to the A's should go on the first column, the numbers next to the B's should go on the second column and the numbers next to the C's should go on the third column.

The equations are the following:

16A-4B+C=1

4A-2B+C=0.94

0A+0B+C=1

So the coefficient matrix looks like this:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right][/tex]

Next we have the matrix that has the variables, in this case our variables are the letters A, B and C. So the matrix looks like this:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right][/tex]

and finally the matrix with the answers to the equations, in this case 1, 0.94 and 1:

[tex]\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

so if we put it all together we end up with the following matrix equation:

[tex]\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right][/tex]

e. When inputing the coefficient matrix in our graphing calculator we end up with the following inverse matrix:

[tex]A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right][/tex]

Inputing matrices and calculating their inverses depends on the model of a calculator you are using. You can refer to the user's manual on how to do that.

f. Our matrix equation has the following general form:

AX=B

where:

A=Coefficient matrix

X=Variables matrix

B= Answers matrix

In order to solve this type of equations, we can make use of the inverse of the coefficient matrix to end up with an equation that looks like this:

[tex]X=A^{-1}B[/tex]

Be careful with the order in which you are doing the multiplication, if A and B change places, then the multiplication will not work and you will not get the answer you need. So when solving this equation we get:

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]*\left[\begin{array}{c}1\\\frac{47}{50}\\1\end{array}\right][/tex]

(Notice that I changed 0.94 for the fraction 47/50 you can get this number by dividing 94/100 and simplifying the fraction)

So, in order to do the multiplication, we need to multiply each row of the coefficient matrix by the answer matrix and add the results. Like this:

[tex]\frac{1}{8}*1+(-\frac{1}{4})(\frac{47}{50})+\frac{1}{8}*1[/tex]

[tex]\frac{1}{8}-\frac{47}{200}+\frac{1}{8}=\frac{3}{200}[/tex]

So the first number for the answer matrix is [tex]\frac{3}{200}[/tex]

[tex]\frac{1}{4}*1+(-1)(\frac{47}{50})+\frac{3}{4}*1[/tex]

[tex]\frac{1}{4}-\frac{47}{50}+\frac{3}{4}=\frac{3}{50}[/tex]

So the second number for the answer matrix is [tex]\frac{3}{50}[/tex]

[tex]0*1+0(\frac{47}{50})+1*1[/tex]

[tex]0+0+1=1[/tex]

So the third number for the answer matrix is 1

In the end, the matrix equation has the following answer.

[tex]\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}\frac{3}{200}\\\frac{3}{50}\\1\end{array}\right][/tex]

which means that:

[tex]A=\frac{3}{200}[/tex]

[tex]B=\frac{3}{50}[/tex]

and C=1

so, when substituting these answers in the general form of the equation of the parabola we get:

[tex]Ax^{2}+Bx+C=y[/tex]

[tex]\frac{3}{200}x^{2}+\frac{3}{50}x+1=y[/tex]

For further information, you can go to the following link:

https://brainly.com/question/12628757?referrer=searchResults

Answer:

x²+12x+35

Step-by-step explanation:

in factored form it would just be

(x+7)(x+5)=0

expand this

x²+12x+35=0

Answer:

show us a screenshot or image

or type it out, copy paste

Step-by-step explanation:

Answer:

9+4y ≥ -15

y ≥ -6

Step-by-step explanation:

Nine Increased by the product of a number

9+4y

is greater than or equal to -15

9+4y ≥ -15

Subtract 9 from each side

9-9+4y ≥ -15-9

4y ≥ -24

Divide by 4

4y/4 ≥ -24/4

y ≥ -6

Answer:

Step-by-step explanation:

345ftyfthftyft.plk,k,

Answer:

Hello,

Anwser is C

Step-by-step explanation:

[tex]y=log_9(12x)\\\\9^y=12x\\\\9^x=12y\ inverting \ x \ and \ y \\\\y=\dfrac{9^x}{12} \\[/tex]

answer- u have to subtract the great no. from the smaller one

Given numbers are 5429 and 5907..

To find the difference we should subtract..

5907

- 5429

-------------

478

_______

Answer:

a. 161

b. 181

c. 100

Step-by-step explanation:

a. 162 comes just after 161 (160, 161, 162, 163...)

b. 181 comes just before 182 (180, 181, 182, 183...)

c. 100 is between 99 and 101 (98, 99, 100, 101, 102...)

Answer:

50% growth would would be (1 + .5)^ n

and 100% growth would be (1+ 1)^n

I assume that the answer would be (1+3)^n

for 300% growth the factor would be 3

Step-by-step explanation:

Answer:

Step-by-step explanation:

Because then the value on the other side will be unbounded by the infinity sign while expressing the answers on a number line.

please click thanks and mark brainliest if you like :)

Answer:

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

General Formulas and Concepts:

Pre-Algebra

Algebra I

Algebra II

Calculus

Differentiation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

Derivative Rule [Quotient Rule]:                                                                           [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

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

Step-by-step explanation:

*Note:

You can simply just use the Quotient and Chain Rule to find the derivative instead of using ln.

Step 1: Define

Identify

[tex]\displaystyle y = \sqrt{\frac{x}{2 - x}}[/tex]

Step 2: Rewrite

Step 3: Differentiate

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

Unit: Differentiation

Book: College Calculus 10e

Answer:

a) forming a bell

b) 5

c) 4.7

d) mean

is the correct answer

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Answer:

Step-by-step explanation:

50%

Answer:

50%

Step-by-step explanation:

Win % = wins / total * 100%

Win% = (5/(5 + 5)) * 100

Win% = 5/10 * 100 = 50%

Answer:

a = 8x

if you want to find x also, then x = a/8

Step-by-step explanation:

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

Answer:

5

Step-by-step explanation:

This is a right triangle, so we can use the Pythagorean theorem

a^2+b^2 = c^2

where a and b are the legs and c is the hypotenuse

QR^2 + 12^2 = 13^2

QR^2 +144 =169

QR^2 = 169-144

QR^2 =25

Take the square root of each side

QR = sqrt(25)

QR =5

Answer:

The answer is "0.6368 and 0.773".

Step-by-step explanation:

The manufacturer of organic compounds guarantees that its clients have at least 100 lbs. of solvent in every fluid drum they deliver. [tex]X\ is\ N(101.8, 3.76)\\\\P(X>100) =P(Z> \frac{100-101.8}{3.76}=P(Z>-0.47))[/tex]

For point a:

Therefore the Probability =0.6368  

For point b:

[tex]P(Z\geq \frac{100-101.8}{\sigma})=0.99\\\\P(Z\geq \frac{-1.8}{\sigma})=0.99\\\\1-P(Z< \frac{-1.8}{\sigma})=0.99\\\\P(Z< \frac{-1.8}{\sigma})=0.01\\\\z-value =0.01\\\\area=-2.33\\\\ \frac{-1.8}{\sigma}=-2.33\\\\ \sigma= \frac{-1.8}{-2.33}=0.773[/tex]

Answer:

Casey's monthly charge for making 1,100 minutes of calls is $70.

Step-by-step explanation:

We can write a piecewise function to model the situation.

Since Casey's phone service only charges a monthly fee of $30 for the first 1000 minutes, we can write that for calling t minutes:

[tex]\displaystyle C(t) = 30\text{ if } t\leq 1000[/tex]

In other words, the total cost is only $30 is the total minutes of call is less than  1000 minutes.

However, if the total minutes of calls is greater than 1000, then its $0.40 per minute on top of the 30. Thus:

[tex]\displaystyle C(t) = 30 + 0.4(t-1000)\text{ if } t>1000[/tex]

All together, our piecewise function will be:

[tex]\displaystyle C(t) = \begin{cases} 30 & t\leq 1000 \\ 30 + 0.4(t-1000) & t>1000\end{cases}[/tex]

We want to determine Caseys monthly charge if he makes 1,100 minutes of calls. So, t = 1100. Since 1100 > 1000, we will use the second equation. This yields:

[tex]C(1100)= 30+0.4((1100)-1000)[/tex]

Evaluate:

[tex]\displaystyle C(1100) = 30+0.4(100) = 30+40=\$70[/tex]

Casey's monthly charge for using 1,100 minutes of call is $70.

Answer:

The kelvin (abbreviation K), also called the degree Kelvin (abbreviation, o K), is the SI unit of temperature. One Kelvin is 1/273.16 (3.6609 x 10 -3 ) of the thermodynamic temperature of the triple point of pure water (H 2 O). The ampere (abbreviation, A) is the SI unit of electric current.

Answer:

kelvin is si unit of tempreature

Complete the square in the denominator.

[tex]s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4[/tex]

Rewrite the given transform as

[tex]\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}[/tex]

Now take the inverse transform:

[tex]L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}[/tex]

Answer:

B, D, F

Step-by-step explanation:

In a rational exponent, the numerator is an exponent, and the denominator becomes the index of the root.

[tex]a^{\frac{m}{n}} = \sqrt[n] {a^m}[/tex]

Answer: B, D, F

Answer:

hope u will understand...if u like this answer plz mark as brainlist

Answer:

[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]

The result is indeed another rational expression.

Step-by-step explanation:

We are given the two functions:

[tex]\displaystyle p(x) = \frac{1}{x+1}\text{ and } q(x) = \frac{1}{x-1}[/tex]

And we want to perform the operation:

[tex]\displaystyle p(x) + q(x)[/tex]

And show that the result is another rational expression.

Add:

[tex]\displaystyle = \frac{1}{x+1} + \frac{1}{x-1}[/tex]

To combine the fractions, we will need a common denominator. So, we can multiply the first fraction by (x - 1) and the second by (x + 1):

[tex]\displaystyle = \frac{1}{x+1}\left(\frac{x-1}{x-1}\right) + \frac{1}{x-1}\left(\frac{x+1}{x+1}\right)[/tex]

Simplify:

[tex]=\displaystyle \frac{x-1}{(x+1)(x-1)} + \frac{x+1}{(x+1)(x-1)}[/tex]

Add:

[tex]\displaystyle = \frac{(x-1)+(x+1)}{(x+1)(x-1)}[/tex]

Simplify. Hence:

[tex]\displaystyle p(x) + q(x) = \frac{2x}{(x+1)(x-1)}[/tex]

The result is indeed another rational expression.

Answer:

was assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?

1. In the game of baseball, which is rather boring, the batter stands on home base (Hughes 1).

2. Baseball is rather surprisingly known as "America's Favorite Pastime."

3. Baseball is "a rather boring sport played between two teams of nine players" (Hughes 1).

4. All of these are plagiarism.

The answer tells that only the third choice is NOT a plagiarism. My question is, why is the first choice a plagiarismwas assigned with this problem (the reference text is attached):

Which of the following, if included in a student's paper, would NOT be an example of plagiarism?