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LuckyWell [14K]
3 years ago
9

What would the shape of your arms be if you modeled a quadratic graph? An absolute value graph?

Mathematics
1 answer:
andre [41]3 years ago
3 0

Answer:

1) A "U" shape or curve

2) a "V" or each arm straight out and point diagonally upward.

A quadratic graph involves a single variable with the highest degree being 2, such as x^2, x^2 + 4x + 2, etc. The graphs all have a U type shape, it may be opening down or up, but it will always be a U type shape.

An absolute value graph involves an equation such as y = |x| + 2, y = |x|, or y = |x - 2|, these all have a v shape. You may not include a quadratic term with the absolute value graph, otherwise it changes the shape slightly (the "arms" are curved)

Honestly you can potentially transform either graph into a different shape. For example y = .00000001x^2, while quadratic, is VERY close in shape to a line.

Of course if we maximize the graph and look at a point such as x = 10^10, we will find a great value for y and see that it does indeed curve, it just takes VERY large values of x.

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Justin wants his new treehouse to blend in with the leaves, so he decides to paint it green. He mixes some white paint and green
spayn [35]

Answer:

  • C. neither. the roof and walls are the same shade of green

Step-by-step explanation:

<u>The first batch:</u>

  • Some white and some green

<u>The second batch:</u>

  • Triple amount of white and triple amount of green

As we see the ratio of white to green has not changed

Therefore there won't be any difference in painting

Correct answer choice is C.

8 0
4 years ago
Find the equation of a circle that has a diameter with the endpoints given by the points A(-4,9) and B (- 2, - 3) )
Aleksandr-060686 [28]

The equation of the circle is (x+3)^{2}+(y-3)^{2}=37

Explanation:

Given that the endpoints of the circle are A(-4,9) and B(-2,-3)

We need to determine the equation of the circle.

<u>Center:</u>

The center of the circle can be determined using the midpoint formula,

Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Substituting the coordinates A(-4,9) and B(-2,-3), we get,

Center=(\frac{-4-2}{2},\frac{9-3}{2})

Center=(\frac{-6}{2},\frac{6}{2})

Center=(-3,3)

Thus, the center of the circle is (-3,3)

<u>Radius:</u>

The radius of the circle can be determined using the distance formula,

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Substituting the center (-3,3) and the endpoint (-4,9), we get,

r=\sqrt{\left(-4+3\right)^{2}+\left(9-3\right)^{2}}

r=\sqrt{\left(-1\right)^{2}+\left(6\right)^{2}}

r=\sqrt{1+36}

r=\sqrt{37}

Thus, the radius of the circle is \sqrt{37}

<u>Equation of the circle:</u>

The standard form of the equation of the circle is given by

(x-a)^{2}+(y-b)^{2}=r^{2}

where (a,b) is the center and r is the radius.

Substituting the values, we have,

(x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}

(x+3)^{2}+(y-3)^{2}=37

Thus, the equation of the circle is (x+3)^{2}+(y-3)^{2}=37

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3 years ago
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xeze [42]

How many cups of peaches does he need to make only 1 cup of fruit salad?

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Ken's car gets 33 miles per gallon. One gallon of gas
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