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3 years ago

7

Write the equation g(x) of the transformation of the parent graph f(x) = |x| stretched horizontally by a factor of 3.

2 answers:

Anni [7]3 years ago

4 0

Answer:

g(x) = |1/3x|

Step-by-step explanation:

f(x) = |x|

y = f(Cx) 0 < C < 1 stretches it in the x-direction

g(x) = |1/3x|

Ainat [17]3 years ago

4 0

We can start with: f(x) = |x|

0 < c < 1 means x-direction stretch

y = f(cx)

3 unit horizontal stretch

After: g(x) = |1/3x|

Best of Luck!

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Please help as soon as possible PLEASE

pashok25 [27]

Area of the entire rectangle $=x^2+10x+24$

Solution:

Area of the blue box = length × width

$=x\times x$

Area of the blue box $=x^2$

Area of the pink box = length × width

$=4\times x$

Area of the pink box $=4x$

Area of the green box = length × width

$=6\times x$

Area of the green box $=6x$

Area of the orange box = length × width

$=6\times 4$

Area of the orange box = 24

Total area of the rectangle $=x^2+4x+6x+24$

$=x^2+10x+24$

Area of the entire rectangle $=x^2+10x+24$

4 0

3 years ago

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

3 years ago

Pls help ASAP i don’t have time!!!!!

Ivahew [28]

Answer:

12.5 on left side and right side is 20.5 and 17.5

Step-by-step explanation:

Just put them on the other side that the number is on.

7 0

3 years ago

Suppose the test scores for a college entrance exam are normally distributed with a mean of 450 and a s. d. of 100. a. What is t

svet-max [94.6K]

Answer:

a) 68.26% probability that a student scores between 350 and 550

b) A score of 638(or higher).

c) The 60th percentile of test scores is 475.3.

d) The middle 30% of the test scores is between 411.5 and 488.5.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 450, \sigma = 100$

a. What is the probability that a student scores between 350 and 550?

This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So

X = 550

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{550 - 450}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 350

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{350 - 450}{100}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a student scores between 350 and 550

b. If the upper 3% scholarship, what score must a student receive to get a scholarship?

100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88

$Z = \frac{X - \mu}{\sigma}$

$1.88 = \frac{X - 450}{100}$

$X - 450 = 1.88*100$

$X = 638$

A score of 638(or higher).

c. Find the 60th percentile of the test scores.

X when Z has a pvalue of 0.60. So it is X when Z = 0.253

$Z = \frac{X - \mu}{\sigma}$

$0.253 = \frac{X - 450}{100}$

$X - 450 = 0.253*100$

$X = 475.3$

The 60th percentile of test scores is 475.3.

d. Find the middle 30% of the test scores.

50 - (30/2) = 35th percentile

50 + (30/2) = 65th percentile.

35th percentile:

X when Z has a pvalue of 0.35. So X when Z = -0.385.

$Z = \frac{X - \mu}{\sigma}$

$-0.385 = \frac{X - 450}{100}$

$X - 450 = -0.385*100$

$X = 411.5$

65th percentile:

X when Z has a pvalue of 0.35. So X when Z = 0.385.

$Z = \frac{X - \mu}{\sigma}$

$0.385 = \frac{X - 450}{100}$

$X - 450 = 0.385*100$

$X = 488.5$

The middle 30% of the test scores is between 411.5 and 488.5.

7 0

3 years ago

The highway fuel economy of a 2016 Lexus RX 350 FWD 6-cylinder 3.5-L automatic 5-speed using premium fuel is a normally distribu

Elodia [21]

Answer:

a) Standard error = 0.5

b) 99% Confidence interval: (19.2125,21.7875)

Step-by-step explanation:

We are given the following in the question:

Population mean = 20.50 mpg

Sample standard deviation = 3.00 mpg

Sample size , n = 36

Standard Error =

$=\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = 0.5$

99% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.575$

$20.50 \pm 2.575(\displaystyle\frac{3}{\sqrt{36}})= 20.50 \pm 1.2875 = (19.2125,21.7875)$

4 0

3 years ago