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torisob [31]
2 years ago
11

How would I put this equation into standard form for a parabola?

Mathematics
1 answer:
Bezzdna [24]2 years ago
5 0

Answer:

fffffffffffffffffffffffffffffff

Step-by-step explanation:


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Standard Number 1 5 10 50 100 500 1,000 Roman Numeral 1 V X L с D A) 44 B) 34 C) 31 D) 29​
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Step-by-step explanation:

SEE THE IMAGE FOR SOLUTION

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F(x)=8_-4x-x^3. g(x)=x^2+7x-9. Find f(x)+g(x)
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(8 - 4x -  {x}^{3} ) + ( {x}^{2}  + 7x - 9) \\  =  -  {x}^{3}  +  {x}^{2}  - 4x + 7x + 8 - 9 \\  =  -  {x}^{3}  +  {x}^{2}  + 3x - 1
(8x - 4x -  {x}^{3} ) + ( {x}^{2}  + 7x - 9) \\  -  {x}^{3}  +  {x}^{2}  + 8x - 4x + 7x - 9 \\  -  {x}^{3}  +  {x}^{2}  + 11x - 9
The answer is either B, and you forgot to put in the ^2 in the answer choice, or C, and you forgot to put an x where there is a _ in the question.

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7 0
3 years ago
What are the solutions to the following system? StartLayout Enlarged left-brace 1st row negative 2 x squared + y = negative 5 2n
Yakvenalex [24]

Answer:

(\sqrt{2},-1),(-\sqrt{2},-1)

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

Step-by-step explanation:

we have

-2x^{2} +y=-5 ----> equation A

y=-3x^{2} +5 -----> equation B

solve by substitution

substitute equation B in equation A

-2x^{2} +(-3x^{2} +5)=-5

solve for x

-5x^{2} +5=-5

-5x^{2}=-10

x^{2}=2

x=\pm\sqrt{2}

<em>Find the value of y</em>

y=-3x^{2} +5

For x=\sqrt{2} ----> y=-3(\sqrt{2})^{2} +5=-1

For x=-\sqrt{2} ----> y=-3(-\sqrt{2})^{2} +5=-1

therefore

The solutions are

(\sqrt{2},-1),(-\sqrt{2},-1)

(StartRoot 2 EndRoot, negative 1) and (negative StartRoot 2 EndRoot, negative 1)

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