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Alex73 [517]
2 years ago
11

The vertex of a parabola that opens downward is at (0, 4). The vertex of a second parabola is at (0, –4). If the parabolas inter

sect at two points, which statement must be true?
Mathematics
2 answers:
dmitriy555 [2]2 years ago
5 0

Answer:

d. just got it right on the unit testttttt

Step-by-step explanation:

erma4kov [3.2K]2 years ago
4 0
Many statements can be true about those parabolas, so you need to include the list of choices.

This is the list of answer choices that I found for this same question:

<span>A. The second parabola opens downward.
B. The second parabola opens upward.
C. The points of intersection are on the x-axis.
D. The points of intersection are of equal distance from the y-axis.

While A, B or C may be or may not be true, D has to be true.

D. has to be true because the symetry axis of both parabolas is x = 0, so the intersection points will  be necesarily to the same distance of the x-axis and the y-axis.

Answer: </span>
<span>The points of intersection are of equal distance from the y-axis. </span>
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Feliz [49]

Answer: the answer is c or d, im not exaclty positive but i want to say the answer is d.

Step-by-step explanation:

4 0
3 years ago
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Suppose that you believe that contractor 1 is twice as likely to win as contractor 3 and that contractor 2 is three times as lik
Ray Of Light [21]

Answer:

probability that contractor 1,2 and 3 win is 33%,50% and 17% respectively

Step-by-step explanation:

assuming that there are no other contractors then:

probability that 1 , 2 or 3 win = 1

denoting as X= probability that contractor 3 wins , then assuming that only one wins , we have

probability that 1 , 2 or 3 win = 1

probability that contractor 1 wins + probability that contractor 2 wins + probability that contractor 3 wins = 1

2*P(X) + 3*P(X) + P(X) = 1

6*P(X) = 1

P(X) =1/6

then

-probability that contractor 1 wins = 2/6 (33%)

-probability that contractor 3 wins = 3/6 (50%)

-probability that contractor 3 wins = 1/6 (17%)

3 0
2 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
2 years ago
Consider functions f and g.
abruzzese [7]

Answer:

letter c I am not sure about the answer

7 0
3 years ago
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Find one terminating decimal and one repeating decimal between -1/2 and -1/3
Masteriza [31]
Terminating decimal:
-0.45
Repeating decimal:
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