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Phoenix [80]
3 years ago
7

Write a quadratic equation in standard form with x-intercepts (-3, 0) and (4, 0) that passes through (2, -20).

Mathematics
1 answer:
Jet001 [13]3 years ago
7 0

Answer:

ax^2 + bx +c = 0

(-3,0), (4,0); (2,-20)

=>

9a -3b + c = 0

16a +4b + c = 0

4a +2b + c = -20

=>

a = 2

b = -2

c = -24

=> 2x^2 -2x -24 =0

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Somebody please help me with this question!​
marta [7]

Answers:

P(A) = 7/12

P(B) = 1/2

=====================================================

Explanation:

To see how I calculated P(A), check out this link to this very similar question

brainly.com/question/27669586

--------------------

Now to calculate P(B)

If a number is divisible by 2, then the number is a multiple of 2.

In other words, the number is even.

Counting through the values in the table, you should find that there are 18 sums that are even (2, 4, 6, 8, 10 and 12). Refer to the dice chart below.

Here's a further breakdown

  • 1 copy of "2"
  • 3 copies of "4"
  • 5 copies of "6"
  • 5 copies of "8"
  • 3 copies of "10"
  • 1 copy of 12

Side note: We have nice symmetry going on.

There are 1+3+5+5+3+1 = 18 values total that are even numbers. The other half are odd numbers of course.

P(B) = 18/36 = (1*18)/(2*18) = 1/2

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3 years ago
Help please thanks in advance
DochEvi [55]
What does the black say, the options
3 0
3 years ago
(-1/3) to the power of five?
Bumek [7]

\bf \left( -\cfrac{1}{3} \right)^5\implies \left( -\cfrac{1}{3} \right)\left( -\cfrac{1}{3} \right)\left( -\cfrac{1}{3} \right)\left( -\cfrac{1}{3} \right)\left( -\cfrac{1}{3} \right)\implies -\cfrac{1^5}{3^5}\implies -\cfrac{1}{243}


recall minus * minus * minus * minus * minus is minus.

8 0
4 years ago
Integrate Sec (4x - 1) Tan (4x - 1) Dx ​
Delicious77 [7]

\bf \displaystyle\int~sec(4x-1)tan(4x-1)dx \\\\[-0.35em] ~\dotfill\\\\ sec(4x-1)tan(4x-1)\implies \cfrac{1}{cos(4x-1)}\cdot \cfrac{sin(4x-1)}{cos(4x-1)}\implies \cfrac{sin(4x-1)}{cos^2(4x-1)} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int~\cfrac{sin(4x-1)}{cos^2(4x-1)}dx \\\\[-0.35em] ~\dotfill\\\\ u=cos(4x-1)\implies \cfrac{du}{dx}=-sin(4x-1)\cdot 4\implies \cfrac{du}{-4sin(4x-1)}=dx \\\\[-0.35em] ~\dotfill

\bf \displaystyle\int~\cfrac{~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{u^2}\cdot \cfrac{du}{-4~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies -\cfrac{1}{4}\int\cfrac{1}{u^2}du\implies -\cfrac{1}{4}\int u^{-2}du \\\\\\ -\cfrac{1}{4}\cdot \cfrac{u^{-2+1}}{-1}\implies \cfrac{1}{4}\cdot u^{-1}\implies \cfrac{1}{4u}\implies \cfrac{1}{4cos(4x+1)}+C

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4 years ago
If point Q does not lie on the x-axis or the y-
Harman [31]
E. II or IV only because it will have different signs if point Q was in either of those quadrants
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