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icang [17]
3 years ago
15

What is the y-value of the vertex of the function f(x)=-(x+8)(x-14)? 3 6 112 121

Mathematics
2 answers:
MAXImum [283]3 years ago
6 0

Answer:

121

Step-by-step explanation:

Given

f(x) = - (x + 8)(x - 14) ← expand factors using FOIL

     = - (x² - 6x - 112) ← distribute by - 1

     = - x² + 6x + 112

Given a quadratic in standard form. y = ax² + bx + c : a ≠ 0

Then the x- coordinate of the vertex is

x_{vertex} = - \frac{b}{2a}

f(x) = - x² + 6x + 112 ← is in standard form

with a = - 1 and b = 6, thus

x_{vertex} = - \frac{6}{-2} = 3

Substitute x = 3 into f(x) for corresponding value of y

f(3) = - (3)² + 6(3) + 112 = - 9 + 18 + 112 = 121

vertex = (3, 121 ) ← with y = 121

zysi [14]3 years ago
6 0

Answer:

D. 121

Step-by-step explanation:

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Answer:

Answer in explanation.

Step-by-step explanation:

Lets take

y =  \frac{3}{4} x - 5

as L1.

Take L2 the line perpendicular to L1.

Standard form of equation of line:

y=mx+c, where m = slope and c = y-intercept.

Since L1 and L2 are perpendicular,

mL1 x mL2 = -1

Substitute mL1 into the equation,

3/4 x mL2 = -1

mL2 = -1 ÷ 3/4

mL2 = -4/3

L2 : y = mx+ c

Substitute y = 6, x = 8 and m = -4/3 into the equation,

6 =  -  \frac{4}{3} (8) + c \\ 6 = -  \frac{32}{3}  + c \\ c = 6 +  \frac{32}{3}  \\  = 16 \frac{2}{3}

therefore L2:

y =  -  \frac{4}{3}x   + 16 \frac{2}{3}

Lets take L3 as the line parallel to L1.

Since L3 and L1 are parallel,

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substitute y = 6, x = 8 and m = 3/4 into equation.

6 = ( \frac{3}{4} )(8) + c \\ 6 = 6 + c \\ c = 6 - 6 \\  = 0

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8 0
2 years ago
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adelina 88 [10]

Answer:

<1=39 degrees

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Step-by-step explanation:

Detailed ans is in attachment

hope it helps:)

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4 0
3 years ago
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Step-by-step explanation:

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Answer:

Option (A)

Step-by-step explanation:

From the table attached,

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