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3 years ago

Four expressions are shown:

Mathematics

2 answers:

Neko [114]3 years ago

8 0

C and D is the correct answer.

erma4kov [3.2K]3 years ago

8 0

C and D are equivalent

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Which statements are correct about quadrilaterals?

Sauron [17]

The correct answers for this question would be A), B), and C) or the first, second, and third options.

4 0

3 years ago

If a building was placed at coordinates (-5,5), what would be the new coordinates if rotated 270° counterclockwise around the or

Brums [2.3K]

The rule for a 270° counterclockwise rotation is (x, y) ---> (y, -x).

So if you rotated (-5,5) 270° counterclockwise around the origin the new coordinates would be (-5,-5).

Hope this Helps!!

6 0

3 years ago

Read 2 more answers

Morre Math HARDDDDDDDD

Dafna11 [192]

Answer:

the person above is right it's 4.5 units

give them the crown

7 0

3 years ago

30. 3 x (5 + 6) = 33<br>33. 14 - (8 - 2) - 1 =<br>36. (3 + 2) × (4 + 6) =<br>39. 6 x 4 + 5) =

Vanyuwa [196]

Answer:

39. 29

36. 50

33. 7

30. 33

Step-by-step explanation:

39. According to the Order of Operations, in this case, Subtraction & Addition are evaluated after Division & Multiplication:

$6 \times 4 + 5 = 24 + 5 = 29$

36. Evaluate everything since everything is wrapped in parentheses:

$(3 + 2) \times (4 + 6) = 5 \times 10 = 50$

33. Evaluate everything inside of parentheses FIRST:

$14 - (8 - 2) - 1 = 14 - 6 - 1 = 7$

I am joyous to assist you anytime.

7 0

4 years ago

Simplify

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

4 years ago