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3 years ago

9

Use the distributive property to write 35 + 15 as a sum with no common factors

1 answer:

kolezko [41]3 years ago

7 0

Here's one:

5(3+4)+3(20-5)

Think of different ways that add or multiply up to 35 or 15.

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The range of a relation is

Kruka [31]

The range of a relation is the possible output values, the y-values.

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3 years ago

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Which of the inequalities represents the statement, "two times a number x, less 15, is greater than or equal to 4 times the numb

Damm [24]

The choices are:

A)
2x - 15 ≤ 4y

B)
2x - 15 ≥ 4y

C)
15 - 2x ≤ 4y

D)
<span>15 - 2x ≥ 4y
</span>
Therefore, the best and most correct answer among the choices provided by the question is the fourth choice "<span>15 - 2x ≥ 4y". </span>I hope my answer has come to your help. God bless and have a nice day ahead!

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Find the length of the base of a square pyramid if the volume is 128 cubic inches and has a height of 6 inches

Len [333]

Answer: a = 8

Step-by-step explanation:

Volume of a square based pyramid is given as

$\frac{a^{2}h }{3}$ .

where a is the length of the side and h is the height.

Substituting into the formula , we have

V = $\frac{a^{2}h }{3}$

128 = $a^{2}h$ /3

384= $a^{2}$ X 6

Divide through by 6

384/6 = $a^{2}$

64 = $a^{2}$

Therefore :

a = $\sqrt{\frac{64}{1} }$

a = 8

6 0

3 years ago

What is 4/5 divided 1/3

vova2212 [387]

The answer would be 2.4, or in fraction form 12/5. Please mark brainliest <3

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13.......................

Alla [95]

Answer:

The answer is (d) ⇒ $pq^{2}r\sqrt[3]{pr^{2}}$

Step-by-step explanation:

* To simplify the cube roots:

If its number then the number must be written in the form x³

then we divide the power by 3 to cancel the radical

If its variable we divide its power by 3 to cancel the radical

∵ $\sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}$

∴ $p^{\frac{4}{3}}q^{2}}r^{\frac{5}{3}}=p^{1\frac{1}{3}}q^{2}r^{1\frac{2}{3}}$

∵ $p^{\frac{1}{3}}=\sqrt[3]{p}$

∵ $r^{\frac{2}{3}}=\sqrt[3]{r^{2}}$

∴ $p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})$

∴ $prq^{2}\sqrt[3]{pr^{2}}}$

∴ The answer is (d)

5 0

3 years ago