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VashaNatasha [74]
3 years ago
14

Plz hurry! answer asap!!

Mathematics
2 answers:
Lena [83]3 years ago
4 0

Answer:

kjvdfvnfjv

Step-by-step explanation:rjfnu4fburbf

Over [174]3 years ago
4 0

Answer:

A, D, E

Step-by-step explanation:

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When you take this problem and simplify it you get 9p-3
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The population of a town is 500,000 in 1990. The population increases at a rate of 5% every five years. What will be the approxi
SVETLANKA909090 [29]
A = P(1+r)^(t/5) A = 500000(1+0.05)^(15/5) A = 500000(1.05)^(15/5) A = 500000(1.05)^3 A = 500000*1.157625 A = 578812.5 Telling us that the population will be about 578,812 people in the year 2005
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3 years ago
Destiny just received two separate gifts from her great-great-grandmother.
qaws [65]

Answer:

6 bags

Step-by-step explanation:

The first gift is a box of 18 chocolate candy bars, and the second gift is a pack of 12 cookies. Destiny wants to use all of the chocolate candy bars and cookies to make identical snack bags for her cousins. What is the greatest number of snack bags that Destiny can make? Answer: 6 bags

8 0
3 years ago
For the graphed exponential equation, calculate the average rate of change from x = −3 to x = 0.
iragen [17]

Answer:

-\frac{7}{3}

Step-by-step explanation:

To solve this, we are using the average rate of change formula:

m=\frac{f(b)-f(a)}{b-a}

where

m is the average rate of change

a is the first point

b is the second point

f(a) is the function evaluated at the first point

f(b) is the function evaluated at the second point

We want to know the average rate of change of the function f(x)=0.5^x-6 form x = -3 to x = 0, so our first point is -3 and our second point is 0. In other words, a=-3 and b=0.

Replacing values

m=\frac{f(b)-f(a)}{b-a}

m=\frac{0.5^0-6-(0.5^{-3}-6)}{0-(-3)}

m=\frac{1-6-(8-6)}{3}

m=\frac{-5-(2)}{3}

m=\frac{-5-2}{3}

m=\frac{-7}{3}

m=-\frac{7}{3}

We can conclude that the average rate of change of the exponential equation form x = -3 to x = 0 is -\frac{7}{3}

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3 years ago
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In your pocket that is where the 1$ is
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3 years ago
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