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Tanya [424]
3 years ago
15

Easy one - giving brainly! show work.(double-checking my answer)​

Mathematics
2 answers:
qaws [65]3 years ago
6 0

Answer:

The child's height is 6 cm

The parent's height is 9 cm

And now the child's height again 1.3 cm

And the parents height is 3.9

Step-by-step explanation: I think it's supposed to be that so yeah.

Ede4ka [16]3 years ago
3 0

Answer:

the height of the parent is 1.95 meters.

Step-by-step explanation:

divide 9 by 6 to get 1.5

multiply 1.5 by 1.3 to get the height of the parent

1.95m

your answer is correct

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Solve the one step inequality.
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x≥9

Step-by-step explanation:

I hope this helps you

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A woman has a total of $ 12 , 000 to invest. She invests part of the money in an account that pays 7 % per year and the rest in
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That would be 34 your welcome:)
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THE GRINCH FOUND THE STRENGTH OF<br>2x6x3) ÷4+1 GRINCH'S PLUS TWO​
anzhelika [568]

Answer:

7 r1

Step-by-step explanation:

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3 0
4 years ago
Evaluate the sum of the following finite geometric series.
rjkz [21]

Answer:

\large\boxed{\dfrac{156}{125}\approx1.2}

Step-by-step explanation:

<h3>Method 1:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}

<h3>Method 2:</h3>

\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}

a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1

\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}

5 0
3 years ago
PLEASE HELP!!!
gregori [183]
Your answer is B. I hope this is helpful!
3 0
3 years ago
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