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mestny [16]
3 years ago
13

Show work and explain with formulas if possible.

Mathematics
2 answers:
Drupady [299]3 years ago
7 0

12 Answer: 20.25

Step-by-step explanation:

\begin{lgathered}a_1=4,\ r=1.5,\ n=5\\\\a_n=a_1\cdot r^{n-1}\\\\a_5=4\cdot (1.5)^{5-1}\\\\.\ =4\cdot (1.5)^4\\\\.\ =4\cdot 5.0625\\\\.\ =\large\boxed{20.25}\end{lgathered}

a

1

=4, r=1.5, n=5

a

n

=a

1

⋅r

n−1

a

5

=4⋅(1.5)

5−1

. =4⋅(1.5)

4

. =4⋅5.0625

. =

20.25

13 Answer: 1980

Step-by-step explanation:

16 Answer: 12

Step-by-step explanation:

\begin{lgathered}\sum\limits^\infty_{n=1} 6\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\\S_1=6\qquad \qquad S_2=3\qquad \qquad S_3=1.5\\\\\\\sum\limits^\infty_{n=1} 6\bigg(\dfrac{2}{2^n}\bigg)=\sum\limits^\infty_{n=1} 6\cdot 2\bigg(\dfrac{1}{2^n}\bigg)=12+\sum\limits^\infty_{n=1}\bigg(\dfrac{1}{2^n}\bigg)=12+0=\large\boxed{12}\end{lgathered}

n=1

∑

∞

6(

2

1

)

n−1

S

1

=6S

2

=3S

3

=1.5

n=1

∑

∞

6(

2

n

2

)=

n=1

∑

∞

6⋅2(

2

n

1

)=12+

n=1

∑

∞

(

2

n

1

)=12+0=

12

german3 years ago
5 0

12 Answer:  20.25

<u>Step-by-step explanation:</u>

a_1=4,\ r=1.5,\ n=5\\\\a_n=a_1\cdot r^{n-1}\\\\a_5=4\cdot (1.5)^{5-1}\\\\.\ =4\cdot (1.5)^4\\\\.\ =4\cdot 5.0625\\\\.\ =\large\boxed{20.25}

13 Answer:  1980

<u>Step-by-step explanation:</u>

a_1=8,\ d=4, \n=30\\\\a_n=a_1+d(n-1)\\\\a_{30}=8+4(30-1)\\\\.\ =8+4(29)\\\\.\ =8+116\\\\.\ =124\\\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{30}=\dfrac{8+124}{2}\cdot 30\\\\\\.\quad =\dfrac{132}{2}\cdot 30\\\\\\.\quad =66\cdot 30\\\\\\.\quad =\large\boxed{1980}

16 Answer:  12

<u>Step-by-step explanation:</u>

\sum\limits^\infty_{n=1} 6\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\\S_1=6\qquad \qquad S_2=3\qquad \qquad S_3=1.5\\\\\\\sum\limits^\infty_{n=1} 6\bigg(\dfrac{2}{2^n}\bigg)=\sum\limits^\infty_{n=1} 6\cdot 2\bigg(\dfrac{1}{2^n}\bigg)=12+\sum\limits^\infty_{n=1}\bigg(\dfrac{1}{2^n}\bigg)=12+0=\large\boxed{12}

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Answer:

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Let t be the number of years after 1999.

From the information given:

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The population growth can be modeled with a linear equation. The initial population was P_0 is 961400 and it grows by 9200 people per year.

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T(t)=P(t)\cdot A(t)\\T(t)=(9200t+961400)\cdot (1400t+30593)

The rate at which the total personal income was rising in the Richmond-Petersburg area is the derivative T(t)'

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If f and g are both differentiable, then:

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Applying the Product Rule

\frac{d}{dt}T(t)=\frac{d}{dt} [(9200t+961400)\cdot (1400t+30593)]\\\\T(t)'=\frac{d}{dt}\left(9200t+961400\right)\left(1400t+30593\right)+\frac{d}{dt}\left(1400t+30593\right)\left(9200t+961400\right)\\\\T(t)'=9200\left(1400t+30593\right)+1400\left(9200t+961400\right)\\\\T(t)'=12880000t+281455600+12880000t+1345960000\\\\T(t)'=25760000t+1627415600

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The raising is

T(0)'=25760000(0)+1627415600\\T(0)'=1,627,415,600

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