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Misha Larkins [42]

3 years ago

13

PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE Please help please

2 answers:

steposvetlana [31]3 years ago

5 0

Answer:

Step-by-step explanation:

The correct answer is: (-1, -2)

this is how i got my answer: by examining the point of intersection of the lines of the 2 graphs show in the we can see that the point of intersection is at (-1, 2) Answer: x = -1, y = 2 ∀∈

umka21 [38]3 years ago

3 0

The correct answer is: "(-1, -2)" .
________________________________________
Note: By examining the point of intersection of the lines of the 2 graphs show in the "image attached", we can see that the point of intersection is at: "(-1, 2"); that is: "x = -1, y = 2" .
________________________________________

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Write the equation of the line, with the given properties, in slope-intercept form.

lukranit [14]

Answer:

y=-4x-25

Step-by-step explanation:

y-y1=m(x-x1)

y-7=-4(x-(-8))

y-7=-4(x+8)

y=-4x-32+7

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What is the radius and diameter of the following circle?

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Answer:

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Step-by-step explanation:

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inn [45]

Answer:

1.3

Step-by-step explanation:

BC is the distance between B and C.

BC = x − (-0.4)

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AB = -0.4 − (-1.25)

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3 years ago

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Find the missing angle

inna [77]

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ANSWER:

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f(x) is a quadratic function with x-intercepts at (−1, 0) and (−3, 0). If the range of f(x) is [−4, [infinity]) and g(x) = 2x^2

iren2701 [21]

Answer with Step-by-step explanation:

We are given that function f(x) which is quadratic function.

x -intercept of function f(x) at (-1,0) and (-3,0)

x-Intercept of f means zeroes of f

x=-1 and x=-3

Range of f =[-4, $\infty$ )

g(x)= $2x^2+8x+6=2(x^2+4x+3)$

$g(x)=0$

$2(x^2+4x+3)=0$

$x^2+4x+3=0$

$x^2+3x+x+3=0$

$x(x+3)+1(x+3)=0$

$(x+1)(x+3)=0$

$x+1=0\implies x=-1$

$x+3=0\implies x=-3$

Therefore, x-intercept of g(x) at (-1,0) and (-3,0).

Substitute x=-2

$g(-2)=2(-2)^2+8(-2)+6=8-16+6=-2$

$g(x)=2(x^2+4x)+6$

$g(x)=2(x^2+2\times x\times 2+4-4)+6=2(x^2+2\times x\times 2+4)-8+6$

$g(x)=2(x+2)^2-2$

By comparing with the equation of parabola

$y=a(x-h)^2+k$

Where vertex=(h,k)

We get vertex of g(x)=(-2,-2)

Range of g(x)=[-2, $\infty$ )

Zeroes of f and g are same .

But range of f and g are different.

Range of f contains -3 and -4 but range of g does not contain -3 and -4.

f and g are both quadratic functions.

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3 years ago