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3 years ago

Write this trinomial in factored form. 7x^2 - 16x - 15 Enter the correct answer.

Mathematics

1 answer:

Serggg [28]3 years ago

3 0

Hey there! :)

Answer:

(7x +5)(x - 3)

Step-by-step explanation:

Starting with:

7x² - 16 - 15

Find two numbers that add up to -16 and multiply to form -15. Remember the coefficient of 7 in the front of the equation.

The numbers '5' and '-3' can be derived. Put these into factored form:

(7x +5)(x - 3)

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F(x) = –x^2+ 6<br> Find f(-3)

Lynna [10]

So if f(-3), then x = -3.

Substitute.

y = -(-3)^2 + 6

= -(9) + 6

= -3

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3 years ago

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Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.

Blizzard [7]

Answer:

The answer is

$y = 3$

$y = - 4$

Step-by-step explanation:

We must find a solution where

$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

$\frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1}$

Which equals

$\frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)}$

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$\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Simplify the right side by multiplying the fraction

$\frac{6y}{(y + 1)(y + 2)}$

Set both fractions equal to each other

$\frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}$

Since the denomiator are equal, we must set the numerator equal to each other

$6y = {y}^{2} + 7y - 12$

$= {y}^{2} + y - 12$

$(y + 4)(y - 3)$

$y = - 4$

$y = 3$

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3 years ago

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nirvana33 [79]

Hello

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4 years ago

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horsena [70]

Answer:

third option

Step-by-step explanation:

$\frac{16 {x}^{2} }{4x + 24} \div \frac{8 {x}^{2} - 40x}{ {x}^{2} - 36} \\ = \frac{16 {x}^{2} }{4(x + 6)} \times \frac{(x + 6)(x - 6)}{8x(x - 5)} \\ = \frac{16 {x}^{2} }{4} \times \frac{x - 6}{8x(x - 5)} \\ = \frac{x}{1} \times \frac{x - 6}{2(x - 5)} \\ = \frac{x(x + 6)}{2x - 10}$

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