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3 years ago

14

What is the product of seven and eight decreased by “x” if x =15?

2 answers:

jekas [21]3 years ago

7 0

Answer:

0

Step-by-step explanation:

product is another word for +

decreased is another word for -

____________________________

(7 + 8) - x

so

(7+8) - 15

15 - 15

= 0

yKpoI14uk [10]3 years ago

3 0

Answer:

41

Step-by-step explanation:

<em>Vocab:</em>

Product: Multiplication

Decrease: Subtraction

(7*8)-x

Substitute:

(7*8)-15

Multiply:

(56)-15

41

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Answer:

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Step-by-step explanation:

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3 years ago

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4 years ago

Whats the roots of the following three problems:

klemol [59]

QUESTION 1

The given function is

$f(x)=\frac{(x-3)(x-1)}{(x-1)(x+2)}$

We simplify to get;

$f(x)=\frac{x-3}{x+2}$

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QUESTION 2

The given function is

$f(x)=\frac{5x^2-10x+5}{2x^2-5x+3}$

$f(x)=\frac{5(x^2-2x+1)}{2x^2-5x+3}$

We factor to get;

$f(x)=\frac{5(x-1)^2}{(2x-3)(x-1)}$

$f(x)=\frac{5(x-1)}{(2x-3)}$

This function equals zero when

$5(x-1)=0$

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QUESTION 3

The given function is

$f(x)=\frac{x^3-8}{x^2-6x+8}$

We factor to get,

$f(x)=\frac{(x-2)(x^2+2x+4)}{(x-2)(x+4)}$

$f(x)=\frac{x^2+2x+4}{x+4}$

The function equals zero when $x^2+2x+4=0$

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$D=2^2-4(1)(4)$

$D=-12$

Hence the equation has no real roots.

The complex roots are

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4 0

3 years ago