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Allisa [31]
4 years ago
15

Can anyone pls solve this..is the answer 0

Mathematics
1 answer:
mrs_skeptik [129]4 years ago
4 0

Answer:

0

Step-by-step explanation:

Simplify the following:

81^(1/4) - 8 216^(1/3) + 15 32^(1/5) + sqrt(225)

Hint: | Simplify radicals.

216^(1/3) = (6^3)^(1/3) = 6:

81^(1/4) - 86 + 15 32^(1/5) + sqrt(225)

Hint: | Simplify radicals.

32^(1/5) = (2^5)^(1/5) = 2:

81^(1/4) - 8×6 + 15×2 + sqrt(225)

Hint: | Simplify radicals.

81^(1/4) = (3^4)^(1/4) = 3:

3 - 8×6 + 15×2 + sqrt(225)

Hint: | Multiply -8 and 6 together.

-8×6 = -48:

3 + -48 + 15×2 + sqrt(225)

Hint: | Multiply 15 and 2 together.

15×2 = 30:

3 - 48 + 30 + sqrt(225)

Hint: | Look for the difference of two identical terms.

(-48 + 30 + sqrt(225)) + 3 = 0:

Answer: 0

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Step-by-step explanation:

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3 years ago
Which expression is equivalent to the expression below? StartFraction 6 c squared + 3 c Over negative 4 c + 2 EndFraction divide
Triss [41]

Answer:

\frac{-3c(2c + 1)}{2(2c - 1)}

Step-by-step explanation:

We want to find the fraction that is equivalent to the fraction below:

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Let us put the fraction above in a simpler form:

We can factorise the numerator with 3c and the denominator with -2.

\frac{3c(2c + 1)}{-2(2c - 1)}\\\\= \frac{-3c(2c + 1)}{2(2c - 1)}

Therefore, we see that \frac{-3c(2c + 1)}{2(2c - 1)} is equivalent to \frac{6c^2 + 3c}{-4c +2}

5 0
3 years ago
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What is 2 2/5×2 7/9 in simplest mixed number form?
Scorpion4ik [409]

Answer:

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Step-by-step explanation:

2 2/5 * 2 7/9

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Rewriting

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7 0
4 years ago
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Step-by-step explanation:

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About 68% of outside diameters lie between 13.9 inches and 14.1 inches.

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Show your work! I really need the answer with the work!
anzhelika [568]
The surface area is 216 cm sq.

Square base formula: length * width

One square base is 70 cm sq (7*10), another is 56 cm sq (7*8), and the final base is 42 cm sq (7*6). 

The height of both triangular faces is 4.8 cm

The two triangular faces are 24 cm sq each, 48 cm sq when added together.

Add all 5 numbers up to get 216.
3 0
3 years ago
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