I need help with 2 alegbra questions. Get 10 easy points!

Mathematics

1 answer:

gizmo_the_mogwai [7]4 years ago

8 0

For 5, we if the y-vlaues are increasing by a common ratio.

(y at x = 1) / (y at x = 0) is = 8 / 4 = 2
(y at x = 2) / (y at x = 1) is 24 / 8 = 3
(y at x = 3) / (y at x = 2) is 48/24 = 2

no, exponential function cannot model this data. you can see that the what you have to multiply to the previous changes.

exponential function has the form y = a(b)^x. if that "b" value kept changing, it's not exponential. like y = 4(2)^x would be exponential only for the first two values since y = 4(2)^x at x = 0 is 4 and y = 4(2)^x at x = 1 is 8, but it is not 24 at x = 2 since y = 4(2)^2 = 16.

6.

exponential function has the form y = a(b)^(x)
we see at x = 0, y = 1 so

1 = a(b)^0
1 = a(1)
a = 1

since a=1, we have y = 1(b)^x. to find b, use other points

since at x = 1 we have y = 3, then

3 = (b)^1
b = 3

so the function is y = 3^x

You might be interested in

h(x)= \dfrac{1}{8} x^3 - x^2h(x)= 8 1 x 3 −x 2 h, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided b

Nesterboy [21]

Answer:

$\dfrac{1}{2}$

Step-by-step explanation:

Given the function $h(x)=\dfrac{1}{8}x^3-x^2$

The average rate of change of the function h(x) over the interval $a\le x\le b$ can be calculated using formula

$\dfrac{h(b)-h(a)}{b-a}$

In your case,

$a=-2,\\ \\b=2,$

so the average rate of change is

$\dfrac{h(2)-h(-2)}{2-(-2)}\\ \\=\dfrac{(\frac{1}{8}\cdot 2^3-2^2)-(\frac{1}{8}\cdot (-2)^3-(-2)^2)}{4}\\ \\=\dfrac{(1-4)-(-1-4)}{4}\\ \\=\dfrac{-3+5}{4}\\ \\=\dfrac{2}{4}\\ \\=\dfrac{1}{2}$

6 0

3 years ago

Multiple Choice What is the sum of five-sixths + start fraction 2 over 3 end fraction? A. one and one-third B. One and one half

polet [3.4K]

Answer:

Step-by-step explanation:

$\frac{5}{6} +\frac{2}{3} \\\\=\frac{5}{6}+\frac{2*2}{2*3} \\\\=\frac{5}{6}+\frac{4}{6} \\\\=\frac{5+4}{6} \\\\=\frac{9}{6} \\\\=\frac{3*3}{3*2} \\\\=\frac{3}{2}$