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3 years ago

9

4x^-y^2 what and how would you solve this particular problem

1 answer:

dimulka [17.4K]3 years ago

5 0

It is x^4 - y^2

so, expanding it, (x^2-y)(x^2+y)

hope it served

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2 years ago

X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

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2 years ago

What is the domain for f(x) =26-2x?

anastassius [24]

Domain of f = R .............

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3 years ago

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZ HELP ME I BEGGING YOU PLZZZZZZZZZZZZZZZ HELP!!!! (55 POINTS)

Karo-lina-s [1.5K]

Answer:

1. 74 because 8(10) - 6

2.4(4) - 5(3) = 1

3. 7(3)+8(4) * 2 = 106

4. A

5.a

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7.26

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10a

Step-by-step explanation:

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3 years ago

-2(u+3)-5u<br><br> use distributive property to simplify

gayaneshka [121]

The answer is
-2(u+3)-5u
-2u-6-5u
-3u-6

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3 years ago

Read 2 more answers