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Marysya12 [62]
2 years ago
6

Factor the following expression. Simplify your answer. 5t( -2t + 3)^2/3 +2( -2t + 3)^5/3

Mathematics
1 answer:
4vir4ik [10]2 years ago
7 0

Answer:

(t+6)(-2t+3)^{\frac{2}{3}}

Step-by-step explanation:

\textsf{Given expression}:

5t(-2t+3)^{\frac{2}{3}}+2(-2t+3)^{\frac{5}{3}}

\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c:

\implies 5t(-2t+3)^{\frac{2}{3}}+2(-2t+3)^{\frac{3+2}{3}}

\implies 5t(-2t+3)^{\frac{2}{3}}+2(-2t+3)^{\frac{3}{3}}(-2t+3)^{\frac{2}{3}}

\implies 5t(-2t+3)^{\frac{2}{3}}+2(-2t+3)(-2t+3)^{\frac{2}{3}}

\textsf{Factor out common term }(-2t+3)^{\frac{2}{3}}:

\implies (-2t+3)^{\frac{2}{3}}(5t+2(-2t+3))

\textsf{Simplify}:

\implies (-2t+3)^{\frac{2}{3}}(5t-4t+6)

\implies (-2t+3)^{\frac{2}{3}}(t+6)

\implies (t+6)(-2t+3)^{\frac{2}{3}}

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2 years ago
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Vilka [71]

Answer:

Two solutions were found :

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Step-by-step explanation:

Step  1  :

Equation at the end of step  1  :

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Step  2  :

Theory - Roots of a product :

2.1    A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

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Add  1  to both sides of the equation :

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Divide both sides of the equation by 2:

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Solving a Single Variable Equation :

2.3      Solve  :    x+4 = 0

Subtract  4  from both sides of the equation :

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Two solutions were found :

x = -4

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Processing ends successfully

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nikitadnepr [17]

Answer:

Cos P = 12/37

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Since this is a right triangle, we can use trig functions

Cos theta = adj / hyp

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