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2 years ago

Which function is a shrink of the exponential growth function shown on the graph?.

Mathematics

1 answer:

ale4655 [162]2 years ago

3 0

Step-by-step explanation:

An exponential function is represented by function,

$y = a \times {b}^{x}$

Where a is the y intercept and b is the growth factor.

The graph passes through (0,1) which is our y intercept so a=1

Know let find b.

The graph pass through (1,3) so plug in 1 for x and 3 for y.

$3 = a \times 3 {}^{1}$

$a = 1$

So our base function is

$y = 1 \times 3 {}^{x}$

In order for this function to shrink, we must shift the y intercept such that it is less than 1 but greater than -1.

So the only option is

$y = \frac{1}{2} (3) {}^{x}$

Because 1/2<1

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Rewrite as a simplified fraction.

Mnenie [13.5K]

Answer:

$\bold{x = \dfrac{8}{9}}$

Step-by-step explanation:

We have to rewrite the below as simplified fraction.

$\large{0.\overline{8} = {?}}$

Solution:

Let $x =0.\overline 8$ which means $x =0.888888.......$

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Now, let us subtract (1) from (2):

$10x-x=8.\overline8 - 0.\overline8\\\Rightarrow 9x = 8\\\Rightarrow \bold{x = \dfrac{8}{9}}$

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3 years ago

10 waffles and 4 pancakes cost $14. Five waffles and eight pancakes cost

Stels [109]

9514 1404 393

Answer:

pancake: $0.50
waffle: $1.20

Step-by-step explanation:

You can model the bills using w and p for the price of a waffle and pancake, respectively.

10w + 4p = 14

5w +8p = 10

Subtract half the first equation from the second:

(5w +8p) -(1/2)(10w +4p) = (10) -(1/2)(14)

6p = 3 . . . . . . . simplify

p = 0.50 . . . . . divide by 6

10w + 4(0.50) = 14 . . . . substitute into the first equation

10w = 12 . . . . . . . . . . . . subtract 2

w = 1.20 . . . . . . . . . . . . .divide by 10

A waffle costs $1.20; a pancake costs $0.50.

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3 years ago

Find the area of the surface generated by revolving x=t + sqrt 2, y= (t^2)/2 + sqrt 2t+1, -sqrt 2 <= t <= sqrt about the y

EleoNora [17]

The area is given by the integral

$\displaystyle A=2\pi\int_Cx(t)\,\mathrm ds$

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$\mathrm ds=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

We have

$x(t)=t+\sqrt 2\implies\dfrac{\mathrm dx}{\mathrm dt}=1$

$y(t)=\dfrac{t^2}2+\sqrt 2\,t+1\implies\dfrac{\mathrm dy}{\mathrm dt}=t+\sqrt 2$

$\implies\mathrm ds=\sqrt{1^2+(t+\sqrt2)^2}\,\mathrm dt=\sqrt{t^2+2\sqrt2\,t+3}\,\mathrm dt$

So the area is

$\displaystyle A=2\pi\int_{-\sqrt2}^{\sqrt2}(t+\sqrt 2)\sqrt{t^2+2\sqrt 2\,t+3}\,\mathrm dt$

Substitute $u=t^2+2\sqrt2\,t+3$ and $\mathrm du=(2t+2\sqrt 2)\,\mathrm dt$ :

$\displaystyle A=\pi\int_1^9\sqrt u\,\mathrm du=\frac{2\pi}3u^{3/2}\bigg|_1^9=\frac{52\pi}3$

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