If you horizontally stretch the linear parent function, f(x) = x, by a factor of 3, what is the equation of the new function?

Solve the following linear system of equations using substitution. <br> Y=3x-5<br> Y=2x-6

Anettt [7]

3x-5=2x-6
Move 2x to the same side as 3x.
3x-2x = x
X-5=-6
add 5 to the -6.
X = 5-6
X=-1

7 0

3 years ago

Read 2 more answers

The answers x^1 or x^0<br> How??

pickupchik [31]

Remember a^(m) x a^(n) = a(m+n) & √a = a^(1/2)

===> x^(1/2 -3/2)+x(1/2 -1/2) ====> x^(-1)+x^(0) (any number Exp 0 =1)
X^(-1) +1 ===> 1/x +1 or (1+x)/x

6 0

3 years ago

Heather drove a constant rate. She traveled 162 miles in 3 hours. How far In miles did heather travel in 1 hour?

creativ13 [48]

Answer:

54

Step-by-step explanation:

Divide 162 by 3 and find the constant rate per hour.

8 0

3 years ago

A right circular cylinder has a height of 22 1/4 ft and a diameter 2 2/5 times its height. What is the volume of the cylinder? E

Triss [41]

Answer:

The volume of cylinder is:

49806.06 ft^3

Step-by-step explanation:

We are given a right circular cylinder with:

Height(h)= $22\frac{1}{4}ft=\dfrac{89}{4}ft$

and a diameter is $2\frac{2}{5}$ times it's height.

i.e. $\dfrac{12}{5}$ times it height.

i.e. the diameter is given as:

$\dfrac{12}{5}\times \dfrac{89}{4}=\dfrac{267}{5}ft.$

We know that the radius of cylinder is half the diameter.

Hence, radius(r) of cylinder is:

$\dfrac{1}{2}\times \dfrac{267}{5}\\\\=\dfrac{267}{10}$

The volume(V) of cylinder is given by:

$V=\pi r^2h$

$V=3.14\times (\dfrac{267}{10})^2\times \dfrac{89}{4}\\\\V=\dfrac{19922423.94}{400}\\\\V=49806.06ft^3$

Hence, the volume of cylinder is:

49806.06 ft^3

7 0

3 years ago

4x2 + 4x - 1, evaluate and fully simplify each of the - For the function f(0) following f(s + h) f(x + h) - f(5) h 10

Andrew [12]

Given the function f(x);

$f(x)=-4x^2+4x-1$

Evaluating the function f(x+h);

$\begin{gathered} f(x+h)=-4(x+h)^2+4(x+h)-1 \\ f(x+h)=-4(x^2+2xh+h^2)^{}+4(x+h)-1 \\ f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1 \end{gathered}$

So;

$f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1$

Evaluating the second function;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1-(-4x^2+4x-1)}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1+4x^2-4x+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2+4x^2-4h^2-8xh^{}+4x-4x+4h-1+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h} \end{gathered}$

simplifying further;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h}=-4h-8x+4 \\ \frac{f(x+h)-f(x)}{h}=-4h-8x+4 \end{gathered}$

Therefore, we have;

$undefined$

7 0

1 year ago