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2 years ago

The function f(x)= (x-8) (x+2) (x+3) has which zeros

2 answers:

7 0

$(x-8)(x+2)(x+3) =0\\\\\implies x -8 =0~~~ \text{or}~~~x+2 =0 ~~~ \text{or}~~~x+3 = 0\\\\\implies x = 8 ~~~~~~~~ \text{or}~~~~~~x = -2 ~~~ \text{or}~~~ x = -3\\\\\\\text{Hence the roots are}~ 8, -2 ~\text{and} ~-3$

ArbitrLikvidat [17]2 years ago

4 0

Answer:

(8, 0)

(-2, 0)

(-3, 0)

Step-by-step explanation:

x - 8 = 0

x = 8

x + 2 = 0

x = -2

x + 3 = 0

x = -3

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For each of the figures shown below, the existing coordinates are listed. Follow the instructions for each shape and perform the

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Answer:

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Step-by-step explanation:

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3 years ago

What is x-2y=3 in slope intercept form

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So firstly, subtract x on both sides: $-2y=-x+3$

Next, divide -2 on both sides, and your answer will be y = 1/2x - 3/2 , or the last option.

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4 years ago

Read 2 more answers

Find each function value 1. f(7) if (x)=5x 2. f(4) if f(x)=3x-1

Inessa [10]

ANSWER

$1. \: \: f(7) = 35$

$2. \: f(4) = 11$

EXPLANATION

To find the value of a function, f(x) at a given value x=a, we plug in x=a into f(x) to obtain f(a) in simplified form.

The first function is

$f(x) = 5x$

To find f(7), we substitute x=7, into the function.

$\implies f(7) =5(7) = 5 \times 7 = 35$

The second function is

$f(x) = 3x - 1$

To find f(4), we replace x with 4 to obtain,

$f(4) = 3(4) - 1$

$f(4) = 12 - 1 = 11$

6 0

4 years ago

A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a

sineoko [7]

ANSWER

The scale factor is
$2.5$

EXPLANATION

The given triangle has vertices,

$A(0,0),B(0,4),\:and\:C(6, 0)$ .

The vertices of the image triangle is,

$A'(0,0),B'(0,10),\:and\:C'(15, 0)$ .

The scale factor is given by

$k = \frac{image \: length}{object \: length}$

So we can use any of the corresponding sides to determine the scale factor,

$k = \frac{|A'B'|}{ |AB|}$

$k = \frac{ |10 - 0| }{ |4 - 0|}$

$k = \frac{ |10| }{ |4 |} = \frac{10}{4} = 2.5$

Or

$k = \frac{|A'C'|}{ |AC|}$

$k = \frac{ |15 - 0| }{ |6 - 0|}$

$k = \frac{ |15| }{ |6|} = \frac{15}{6} = 2.5$

Or

$k=\frac{|B'C'|}{|BC|}$

$k = \frac{\sqrt{(15 - 0)^2+(0-10)^2 }}{\sqrt{(6 - 0)^2+(0-4)^2}}$

$k = \frac{ 5\sqrt{13}}{2\sqrt{13}} = \frac{5}{2} = 2.5$

The correct answer is C

3 0

3 years ago

What is the remainder when x^2-5x-22 is divided by x+3?

Zielflug [23.3K]

First, you need to find the zero of x+3, which is -3
then you multiply -3×1, which equals -3
the reason why I multiplied -3×1 is because the 1 is the coefficient of x^2 but we just don't see it because we don't need to.
once we get -3 we add that to the next coefficient (-5)
-3+-5=-8
we take the sum (-8) and multiply that to -3 (the zero of x+3)
we get 24, so we add -22+24=2
-22 Is the next coefficient to come and 2 is our remainder
the new equation is: x-8 remainder 2

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3 years ago