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tankabanditka [31]
3 years ago
8

I WILL MARK BRAINLIEST!!!!!!!!!

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
7 0

Answer:

It is not

Step-by-step explanation:

It is not because this is an exponential function

y=x^2

1^2=1

2^2=4

3^2=9

4^2=16

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Pls, help Which of the following is true?
fgiga [73]

Answer:

C

Step-by-step explanation:

Its not A because The left side 6 is not less than the right side 5, which means that the given statement is false.

Its not B because The left side 6 is not less than the right side 5, which means that the given statement is false.

Its C because The left side 5 is less than the right side 6, which means that the given statement is always true.

Its not D because The left side is 5 is not less than the right side 6, which means that the given statement is false.

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3 years ago
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$11.00 $6.50 that's is the answer
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Jill is pushing a box across the floor. Which represents the upward force perpendicular to the floor?
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What is the equation for the line?​
Alex787 [66]
Heres the answer you’re looking for.

y= -2x-4
7 0
3 years ago
The focus of a parabola is (-3,-5) The directrix of the parabola is y=2
cricket20 [7]

Check the picture below.

so the focus point is there, and the directrix is above it, meaning is a vertical parabola and is opening downwards, since the parabola opens up towards the focus.

now, the vertex is half-way between those two guys, at a "p" distance from either one, if we move over the y-axis from -5 to +2, we have 7 units, half-way is 3.5 units, and that puts us at -1.5 or -1½, as you see in the picture, so the vertex is then at (-3 , -1½).

so the distance from the vertex to the focus point  is then 3½ units, however since the parabola is opening downwards, "p" is negative, thus "p = 3½".

\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{using this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \begin{cases} h=-3\\ k=-\frac{3}{2}\\[0.7em] p=-\frac{7}{2} \end{cases}\implies 4\left( -\cfrac{7}{2} \right)\left[ y-\left(-\cfrac{3}{2} \right) \right]=\left[ x-\left( -3 \right) \right]^2 \\\\\\ -14\left( y+\cfrac{3}{2} \right)=(x+3)^2\implies y+\cfrac{3}{2} =-\cfrac{(x+3)^2}{14} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{14}(x+3)^2-\cfrac{3}{2}~\hfill

4 0
3 years ago
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