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Digiron [165]
3 years ago
7

Given the function h(x) = 4^x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Mathematics
1 answer:
Aleksandr [31]3 years ago
5 0
A)  average rate=Δy/Δx=(y2-y1)/(x2-x1)=(h2-h1)/(x2-x1)
from x1 = 0 to x2= 1, <span>h(x) = 4^x, h(0)=4^0=1, h(1)=4^1=4
</span>average rate=(h2-h1)/(x2-x1)=(4-1)/(1-0)=3  (section A for interval x=0 to x=1)
from x1 = 2 to x2= 3, h(x) = 4^x, h(0)=4^2=16, h(1)=4^3=64
average rate=(h2-h1)/(x2-x1)=(64-16)/(3-2)=48  (section B for interval x=2 to x=3)
B) av.rate section B/ av. rate section A=48/4=12
this is exponential function, and for Δx=1, Δy change more  for bigger numbers

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PLEASE HELP 30 POINTS!!!!!!!!!!!!!!!!!!!!!!!
Anettt [7]

Since you know the value of "x", you can plug in the value for "x" in the equation.

[When an exponent is negative, you move it to the other side of the fraction to make the exponent positive.]

For example:

x^{-2}  or  \frac{x^{-2}}{1} =\frac{1}{x^2}

\frac{1}{y^{-3}} =\frac{y^3}{1}  or  y³


x = -2

f(x) = 9x + 7

f(-2) = 9(-2) + 7 = -18 + 7 = -11


g(x)=5^x

g(-2)=5^{-2}=\frac{1}{5^2}=\frac{1}{25}   (idk if you should have it as a decimal or a fraction)



x = -1

f(x) = 9x + 7

f(-1) = 9(-1) + 7 = -9 + 7 = -2


g(x)=5^x

g(-1)=5^{-1}=\frac{1}{5}



x = 0

f(x) = 9x + 7

f(0) = 9(0) + 7 = 7


g(x)=5^x

g(0)=5^0=1



x = 1

f(x) = 9x + 7

f(1) = 9(1) + 7 = 9 + 7 = 16


g(x)=5^x

g(1)=5^1=5



x = 2

f(x) = 9x + 7

f(2) = 9(2) + 7 = 18 + 7 = 25


g(x)=5^x

g(2)=5^2=25



You need to determine the solution of f(x) = g(x)

Since you know f(x) = 9x + 7 and g(x)=5^x, you can plug in (9x + 7) for f(x), and (5^x) into g(x)


f(x) = g(x)

9x+7=5^x   You can plug in each value of x into the equation


Your answer is x = 2 because when you plug in 2 for x in the equation, you get 25 = 25

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