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4 years ago

Compute each sum below. If applicable, write your answer as a fraction.

Mathematics

1 answer:

Alex777 [14]4 years ago

4 0

Answer:

First equation is -425

Second equation is 11.25

Step-by-step explanation:

First equation we can write as

$\sum_{i=0}^7 5 \times (-2)^i$

computing

When i=0 -> $5 \times (-2)^0 =5$

When i=1 -> $5 \times (-2)^1 = -10$

...

When i=7 -> $5 \times (-2)^7 = --640$

then replacing each term we have

$5 + (-10) + ( 20) + ( -40) + (80) +( -160) + 320 + (-640) = -425$

For the second equation we'll have 9 terms, solving in a similar fashion

When i=1 -> $\frac{1}{4}1=0.25$

When i=2 -> $\frac{1}{4}2=0.50$

When i=3 -> $\frac{1}{4}3=0.75$

...

When i=9 -> $\frac{1}{4}9=2.25$

So we have 0.25 + 0.50 + 0.75 + 1.00 + 1.25 + 1.50+ 1.75 +2.00 +2.25

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3 years ago

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3 years ago

X f(x) -2 -1/200 -1 -1/20 0 -1/2 1 -5 2 -50

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Answer:

f(x) = -1/2·10^x

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3 years ago

The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

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Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

$\frac{10000(0.2)^1}{10000(0.2)^0}$

$\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2$

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

8 0

4 years ago

Please help #22 please sos

iragen [17]

1/5( 5/2 x 4/7 -11/7) + 3/28

Multiply the terms (5/2 x 4/7) in bracket then subtract it from 11/14. Result will be 10/7.

Then multiply 1/5 with 10/7. The result will be 2/7.

2/7 + 3/28 = 11/28

T

4 0

3 years ago

Read 2 more answers