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Zanzabum
4 years ago
14

Please give an explanation with you answer, thank you :)

Mathematics
2 answers:
Anastasy [175]4 years ago
8 0
For this case we have the following expression:
 24 ^ (1/3)
 Rewriting we have:
 3 ^ root (24)
 3 ^ root (2 * 2 * 2 * 3)
 3 ^ root (2 ^ 3 * (3))
 Then, by properties of exponents we have:
 2 * (3 ^ root (3))
 Answer:
 
2 * (3 ^ root (3))
 
option 2
Tasya [4]4 years ago
4 0
Answer:
2∛3 which is the second choice

Explanation:
before we begin, remember the following:
√xy = √x * √y
(xy)ᵃ = xᵃ * yᵃ
∛x = x^(1/3)

Now, for the given, we have:
24^ \frac{1}{3}

24 can be rewritten as 8 * 3

Therefore, the given is now:
(8*3)^ \frac{1}{3}

Distributing the power:
8^ \frac{1}{3} * 3^ \frac{1}{3}

We know that:
∛8 = 2

Therefore, the simplest form of the given would be:
8^ \frac{1}{3} * 3^ \frac{1}{3} 
2 * 3^ \frac{1}{3} 
2∛3

Hope this helps :)
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Fine length of BC on the following photo.
MrMuchimi

Answer:

BC=4\sqrt{5}\ units

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ACD

Find the length side AC

Applying the Pythagorean Theorem

AC^2=AD^2+DC^2

substitute the given values

AC^2=16^2+8^2

AC^2=320

AC=\sqrt{320}\ units

simplify

AC=8\sqrt{5}\ units

step 2

In the right triangle ACD

Find the cosine of angle CAD

cos(\angle CAD)=\frac{AD}{AC}

substitute the given values

cos(\angle CAD)=\frac{16}{8\sqrt{5}}

cos(\angle CAD)=\frac{2}{\sqrt{5}} ----> equation A

step 3

In the right triangle ABC

Find the cosine of angle BAC

cos(\angle BAC)=\frac{AC}{AB}

substitute the given values

cos(\angle BAC)=\frac{8\sqrt{5}}{16+x} ----> equation B

step 4

Find the value of x

In this problem

\angle CAD=\angle BAC ----> is the same angle

so

equate equation A and equation B

\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}

solve for x

Multiply in cross

(8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units

DB=4\ units

step 5

Find the length of BC

In the right triangle BCD

Applying the Pythagorean Theorem

BC^2=DC^2+DB^2

substitute the given values

BC^2=8^2+4^2

BC^2=80

BC=\sqrt{80}\ units

simplify

BC=4\sqrt{5}\ units

7 0
3 years ago
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