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Llana [10]
3 years ago
6

Simplify the expression and show work.

Mathematics
1 answer:
nasty-shy [4]3 years ago
4 0

Answer:

72/149

Step-by-step explanation:

Just to clarify, the curly braces mean the same thing as normal parenthesis.

(12 + 5^2) = (12 + 25) = 37

(-7-15)^2 = (-22)^2 = 484

6^3  = 216

(12 + 5^2) - (-7-15)^2 = 37 - 484 = 447

6^3 / ((12 + 5^2) - (-7 - 15)^2) = 216 / 447 = 72/149

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Can somebody please explain and answer this for me? I really don't get it.
kirill115 [55]
Answer: 12a^3
Explanation: You would get the missing side to complete the perimeter by using the pythagorean theorem so 3^2+4^2=c^2 so you would get 9+16=c^2 which would become 25=c^2. That would get you 5 as the square root of 25 is 5 so in total, it would all be 3a^3+4a^3+5a^3 which would equal 12a^3.
(^ are for exponents) I might be wrong but hopefully it helps a bit. :)
6 0
4 years ago
Which symbol correctly compares the two fractions?
dedylja [7]

Answer:

5/12<3/6

Step-by-step explanation:

3/6 is 1/2 and 5/12 is smaller than 1/2

4 0
3 years ago
Read 2 more answers
Solve the equation. 9x−(3x+9)=2x−25 Select the correct choice below and fill in any answer boxes in your choice. A. The solution
nataly862011 [7]

Answer: Choice A, x=-4

Step-by-step explanation:

1. 9x−(3x+9)=2x−25

2. 9x-3x-9=2x-25

3. 6x-9=2x-25

4. 6x-9=2x-25  

   -2x          -2x

5. 4x-9=-25

       +9    +9

6. 4x=-16 (divide both sides by 4)

7. x=-4

7 0
3 years ago
Which of the following are progressions?
professor190 [17]

In mathematics: Arithmetic progression, sequence of numbers such that the difference of any two successive members of the sequence is a constant. Geometric progression, sequence of numbers such that the quotient of any two successive members of the sequence is a constant. so i chose b)Sequences

7 0
3 years ago
Read 2 more answers
Which choice is equivalent to the product below when x&gt;0?
V125BC [204]

Answer:

D

Step-by-step explanation:

Using the rule of radicals

\sqrt{\frac{a}{b} } = \frac{\sqrt{a} }{\sqrt{b} }

\sqrt{a\\} × \sqrt{b} ⇔ \sqrt{ab}

Given

\sqrt{\frac{6}{x} } × \sqrt{\frac{x^2}{24} }

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x^2}{24}

= \frac{\sqrt{6} }{\sqrt{x} } × \frac{x}{2 \sqrt{6\\} }

Cancel \sqrt{6} on numerator/ denominator

= \frac{1}{\sqrt{x} } × \frac{x}{2\\}

= \frac{1}{\sqrt{x} } × \frac{(\sqrt{x})^2 }{2}

Cancel \sqrt{x} on numerator/ denominator, leaving

= \frac{\sqrt{x} }{2} → D

4 0
3 years ago
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