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3 years ago

Erin is 8 years younger than twice Paula’s age. The sum of their ages is 40. How old is Erin

1 answer:

6 0

You do the sum(40) subtracted by 8, the answer from that divided by 2 since Erin's two times younger.
40-8=32
32÷2=16
Then if you want to check it you do the opposite and get the same sum(40) so 16×2=32 and 32+8=40

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What is the value of -2/5 / 3/4?

Sveta_85 [38]

Answer:

-8/15

Step-by-step explanation:

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2 years ago

the smallest angle in a triangle is one-third of the largest angle. the third angle is 10degrees more than the smallest angle. f

Alex787 [66]

A: smallest
c: largest
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a+b+c=180
a=c/3
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solve it
a=34
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3 years ago

Plz help me plz I need your I am struggling so badly I need your help plz help me I’m begging you plz plz help me

mars1129 [50]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Just substitute the given value into anywhere you see the term x:

2 x 4^2 + 12

Follow the PEMDAS rule:

(calculate the exponential first)

4^2 = 16

Now we have:

2 x 16 + 12

(now the multiplication needs to be calculated)

16 x 2 = 32

Now we have:

32 + 12

Add them together to get the answer:

44 <== this is the answer

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3 0

3 years ago

Read 2 more answers

If no (A) = 40%, no (B) = 30%, n (A∩B) = 20% then find n (<br> (A B)<br> using formula.

Vesnalui [34]

Answer:

solution:-We know that for any two finite sets A and B, n(A∪B)=n(A)+n(B)−n(A∩B).

Here, it is given that n(A)=20,n(B)=30 and n(A∪B)=40, therefore,

n(A∪B)=n(A)+n(B)−n(A∩B)

⇒40=20+30−n(A∩B)

⇒40=50−n(A∩B)

⇒n(A∩B)=50−40

⇒n(A∩B)=10

Hence, n(A∩B)=10

Step-by-step explanation:

hope it helps you friend ☺️

8 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=Simplify%3A%20%5Cfrac%7B%205%C3%97%2825%29%5E%7Bn%2B1%7D%20-%2025%20%C3%97%20%285%29%5E%7B2n%7

Katen [24]

$\green{\large\underline{\sf{Solution-}}}$

<u>Given expression is </u>

$\rm :\longmapsto\:\dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} }$

can be rewritten as

$\rm \: = \: \dfrac{5 \times { {(5}^{2} )}^{n + 1} - {5}^{2} \times {5}^{2n} }{5 \times {5}^{2n + 3} - {( {5}^{2} )}^{n + 1} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {( {x}^{m} )}^{n} \: = \: {x}^{mn}}}} \\$

And

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: {x}^{m} \times {x}^{n} = {x}^{m + n} \: }}} \\$

So, using this identity, we

$\rm \: = \: \dfrac{5 \times {5}^{2n + 2} - {5}^{2n + 2} }{{5}^{2n + 3 + 1} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 4} - {5}^{2n + 2} }$

can be further rewritten as

$\rm \: = \: \dfrac{{5}^{2n + 2 + 1} - {5}^{2n + 2} }{{5}^{2n + 2 + 2} - {5}^{2n + 2} }$

$\rm \: = \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2} - 1)}$

$\rm \: = \: \dfrac{4}{25 - 1}$

$\rm \: = \: \dfrac{4}{24}$

$\rm \: = \: \dfrac{1}{6}$

<u>Hence, </u>

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times {25}^{n + 1} - 25 \times {5}^{2n} }{5 \times {5}^{2n + 3} - {25}^{n + 1} } = \frac{1}{6} }}$

4 0

2 years ago