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Ede4ka [16]
2 years ago
7

What are the zeros of y=x^2 +2x -8

Mathematics
2 answers:
lesantik [10]2 years ago
7 0

Answer:

x = 2 and ( -4)

Step-by-step explanation:

We have to find the zero factors of the given equation

y = x² + 2x - 8

for zero factor we will put y = 0

0 = x² + 2x - 8

0 = x² + 4x - 2x - 8

0 = x (x+4) - 2(x + 4)

0 = (x + 4) (x - 2)

So zero factors will be

x + 4 = 0

x = -4

( x-2) = 0

x = 2

Therefore, x = 2 and ( -4) will be the zeros of the given equation.

Igoryamba2 years ago
6 0
Y = x² + 2x - 8

which two numbers add up to 2 and multiply to -8?

-2 and 4

so rewrite the expression using the above;

(x - 2)(x + 4)
x = 2, -4

so the zeros are 2 and -4.

hope this helps, God bless!
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3 years ago
Keith exercises x hours a week. Troy exercises three hours less than Keith and Joe exercises two times more than Troy. Which exp
muminat
If you would like to know which expression represents the amount of time Joe exercises, you can calculate this using the following steps:

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Identify the values that should be written to complete the X diagram.
cestrela7 [59]

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See the attachment for what goes on your X diagram.

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Step-by-step explanation:

The given quadratic is ...

... x² -3x -28 . . . . . a=1, b=-3, c=-28

a) The value at the top of the X diagram is the product a·c = 1·(-28) = -28.

The value at the bottom of the X diagram is the coefficient b = -3.

The values on the sides of the diagram are the factors of -28 that add up to make -3. These are -7 and 4. That is, ...

(-7)·(4) = -28

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b) Since the two values on the sides of the diagram add up to give "b", the value of "b" in the equation can be rewritten as the sum of these two numbers. Doing that, we have ...

... x² -3x -28

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... = x² -7x +4x -28 . . . . . . order does not matter. It could also be x² +4x -7x -28

c) We can group pairs of terms in the rewritten expression and factor each pair.

... = (x² -7x) +(4x -28) . . . . . first pair has a common factor of x; second pair, 4

... = x(x -7) +4(x -7) . . . . . . . these terms now have a common factor: (x -7)

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3 years ago
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\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5

well then therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

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