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2 years ago

Help please!!! Question is in the picture

Mathematics

2 answers:

EastWind [94]2 years ago

8 0

I think it’s dababy

Nataly [62]2 years ago

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I don’t really know where the Porche is

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On a coordinate plane, a parabola opens up. It goes through (negative 2, 4), has a vertex at (0.25, negative 6), and goes throug

Oksana_A [137]

Answer:

The function has two x-intercepts.

The vertex of the function is (one-quarter, negative 6 and one-eighth).

Step-by-step explanation:

Given that the vertex is located at (0.25, -6) and the parabola opens up, the function has two x-intercepts.

f(x) = 2x² - x - 6 has the following coefficients:

a = 2

b = -1

c = -6

x-coordinate of the vertex is:

x = -b/(2a)

x = 1/(2*2) = 1/4

y-coordinate of the vertex is:

f(1/4) = 2(1/4)² - 1/4 - 6 = -6 1/8

3 0

2 years ago

Read 2 more answers

Solve each equation for y. Then find the value of y for each value of x. y+3x=8;x=-2,0,2

damaskus [11]

Y=-3x+8
when x=-2:
y=-3(-2)+8
y=6+8
y=14
when x=0
y=-3(0)+8
y=0+8
y=8
when x=2
y=-3(2)+8
y=-6+8
y=2

6 0

2 years ago

List the elements of the set of all letters in the word "FOUR"

aleksley [76]

Answer:

{F, O, U, R} in the word 'FOUR'

3 0

3 years ago

What is the quotient of (x3 + 6x2 + 11x + 6) ÷ (x2 + 4x + 3)? x + 2 x – 2 x + 10 x + 6

evablogger [386]

Answer:

x + 2

Step-by-step explanation:

If we know one of the factors, we should be able to use long division.

We can divide the trinomial by the binomial and find the other factor.

x + 2

x² + 4x + 3)x³ + 6x² + 11x + 6

x³ + 4x² + 3x

2x² + 8x + 6

2x² + 8x + 6

Thus,

$\dfrac{x^{3} + 6x^2 + 11x + 6} {x^{2} + 4x + 3} = \mathbf{x + 2}$

6 0

3 years ago

What is the equation in point-slope form of a line that passes through the points (−4, −1) and (5, 7) ?

Sergio039 [100]

The point-slope form:

$y-y_1=m(x-x_1)$

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (-4, -1) and (5, 7). Substitute:

$m=\dfrac{7-(-1)}{5-(-4)}=\dfrac{7+1}{5+4}=\dfrac{8}{9}\\\\\boxed{y-7=\dfrac{8}{9}(x-5)}$

6 0

2 years ago

Help please!!! Question is in the picture​

Help please!!! Question is in the picture