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Vera_Pavlovna [14]

2 years ago

14

1. A store priced a textbook at $126.75 using a markup of 34% of cost.

1 answer:

Alchen [17]2 years ago

6 0

$43.09 of book
Shhebeenebbemansjdx

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WHAT IS? <br> -5(2x+14) = -2(x +63)

Rus_ich [418]

Pretty sure x=-24.5 did the math. Check my answer before you submit anything

3 0

3 years ago

Read 2 more answers

Expertsss/aceee plzzz helpppp quickkklyyyy

Law Incorporation [45]

I would think that it would be A.... I am an expert but, only in 8th, so I would double check.

8 0

2 years ago

Read 2 more answers

Diano4ka-milaya [45]

Answer:

$\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}$

Step-by-step explanation:

$16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}$

$=2\left(\dfrac{1}{a^3}\right)=\dfrac{2}{a^3}$

5 0

3 years ago

6. The angle relationship shown is-

expeople1 [14]

Answer:

60+30=90 degrees = complementary angles.

Step-by-step explanation:

4 0

2 years ago

How to solve??????????HELP

nadezda [96]

I presume you're talking about number 3?

Solution

I = prt (I being the interest)

I =

$5000 \times 5 \times.05$

I = 1,250$

3 0

2 years ago