Ask question

Ask question

All categories

3 years ago

5

I need to know Simplify 2a + 8(b + a) is

1 answer:

KonstantinChe [14]3 years ago

4 0

HELLO !

Okay so what you do is<em> eliminate the parenthesis </em>in the equation because it really helps

So you would multiply 8 and b
Also multiple 8 and a

You would get 8b and 8a

Your equation would look like ...

2a + 8b + 8a = ...

<em>Now you add the alike things in the equation </em>

8a and 2a are the alike things.

Now you would have ...

10a + 8b = ?

You might be interested in

Hard problem plz help

Dafna11 [192]

False.....................................

4 0

3 years ago

Read 2 more answers

George is 4 times Dorothy's age. If George is 56, how old is Dorothy?

yarga [219]

$56 \div 4 = 14$
Dorothy is 14 years old

good luck

7 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%28-81%29x%5E%7B2%7D%20%7D" id="TexFormula1" title="\sqrt{(-81)x^{2} }" alt="\sqrt{(

SSSSS [86.1K]

Answer:

We conclude that:

$\sqrt{\left(-81\right)x^2}=9ix$

Step-by-step explanation:

Given the radical expression

$\sqrt{\left(-81\right)x^2}$

simplifying the expression

$\sqrt{\left(-81\right)x^2}$

Remove parentheses: (-a) = -a

$\sqrt{\left(-81\right)x^2}=\sqrt{-81x^2}$

Apply radical rule: $\sqrt{-a}=\sqrt{-1}\sqrt{a},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=\sqrt{-1}\sqrt{81x^2}$

Apply imaginary number rule: $\sqrt{-1}=i$

$=i\sqrt{81x^2}$

Apply radical rule: $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

$=\sqrt{81}i\sqrt{x^2}$

$=9i\sqrt{x^2}$

Apply radical rule: $\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$=9ix$

Therefore, we conclude that:

$\sqrt{\left(-81\right)x^2}=9ix$

7 0

3 years ago

What is -1/2 times 1/4-3/8?

pav-90 [236]

Answer:

The answer is <em>-0.5</em>

Step-by-step explanation:

6 0

3 years ago

Write the following proportion: 8 is to 64 as 2 is to x.

Ksenya-84 [330]

x = 4

8 multiplied by itself is 64.
And so, 2 times itself is 4.

3 0

3 years ago

Read 2 more answers