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trapecia [35]
2 years ago
6

Solve x^3 = 1 over 8. 1 over 2 ±1 over 2 1 over 4 ±1 over 4

Mathematics
2 answers:
lara [203]2 years ago
8 0
<h3>A N S W E R :</h3>

x³ = 1/8

8 can be written as 2³

x³ = 1/2³

(x)³ = (1/2)³

When exponents are same bases are taken and exponents are eliminated

x = 1/2

  • <u>Hence, the required answer is 1/2</u>

Oduvanchick [21]2 years ago
7 0

Answer:

x = 1/2

Step-by-step explanation:

x^3 = 1/8

Take the cube root of each side

x^3 ^ (1/3) = (1/8)^ (1/3)

Rewriting 8 as 2^3

x^3 ^ (1/3) = (1/2^3)^ (1/3)

x = 1/2

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Answer:

‹A = 141°

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Step-by-step explanation:

Because you are only given sides you can find an individual angle measures with the inverse law of cosines.

Remember the side opposite of the angle corresponds to that angle.

a² = b² + c² - 2bc cos(A) →

A = cos⁻¹ (a² - b² - c² / -2bc)

b² = a² + c² - 2ac cos(B) →

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2 years ago
a square painting has an area of 81x^2-90x-25. A second square painting has an area of 25x^2+30x+9. What is an expression that r
galina1969 [7]

Answer:

The answer in the procedure

Step-by-step explanation:

Let

A1 ------> the area of the first square painting

A2 ---->  the area of the second square painting

D -----> the difference of the areas

we have

A1=81x^{2}-90x-25

A2=25x^{2}+30x+9

case 1) The area of the second square painting is greater than the area of the first square painting

The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting

D=A2-A1

D=(25x^{2}+30x+9)-(81x^{2}-90x-25)

D=(-56x^{2}+120x+34)

case 2) The area of the first square painting is greater than the area of the second square painting

The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting

D=A1-A2

D=(81x^{2}-90x-25)-(25x^{2}+30x+9)

D=(56x^{2}-120x-34)

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