All categories

3 years ago

Let f(x) = 7^x and g(x) = 7^x+2 - 4 which transformations are needed to strand form the graph of f(x) to the graph of g(x)?

Mathematics

1 answer:

serg [7]3 years ago

6 0

Answer:

options 3 and 6

Step-by-step explanation:

the exponent x+2 means that the graph moves to the left 2 units and the value of K, which is -4, means that the graph will then move down 4 units from the parent function f(x)

You might be interested in

Evaluate 9/g+2h+5 when g=3 and h=6

vladimir1956 [14]

To answer this question, simply input the values for g and h:

9/ (3) + 2 (6) + 5 = 9/ 3 + 12 + 5
= 9/ 20

Hope this helps!

6 0

3 years ago

Read 2 more answers

Create your own equation chain using these numbers for the variable (a=10) (b=6) (c=18) (d=3)

Orlov [11]

Answer:

b x 5 = 30

Step-by-step explanation:

This is one of the many correct answers. The reason why is because it shows that the variable is correct. And if you didn’t know what the variable was you could just divide 30 / 5.

7 0

3 years ago

Find the other endpoint of the line segment with the given endpoint and midpoint: endpoint(1.7, -4.6) midpoint (-7.5, 0.3).. (ne

Sergio [31]

Answer:

(- 16.7, 5.2 )

Step-by-step explanation:

let the coordinates of the endpoint be (x , y )

using the midpoint formula

consider the x- coordinate

$\frac{1}{2}$ (x + 1.7) = - 7.5 ( multiply both sides by 2 )

x + 1. 7 = - 7.5 ( subtract 1.7 from both sides )

x = - 16.7

consider the y-coordinate

$\frac{1}{2}$ (y - 4.6 ) = 0.3 ( multiply both sides by 2 )

y - 4.6 = 0.6 ( add 4.6 to both sides )

y = 5.2

endpoint = (- 16.7, 5.2 )

3 0

4 years ago

Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago

Answer two questions about Systems A and B:System A x+3y=−9 2x+y=4 System B 3x+4y=−9 2x+y=4

Jet001 [13]

Answer:

A=(21/5,-22/5) B=(5,-6)

Step-by-step explanation:

6 0

3 years ago