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3 years ago

Write a polynomial equation in standard form with roots of -3 and 4

Mathematics

1 answer:

liubo4ka [24]3 years ago

6 0

Answer:

I don't know the answer

Step-by-step explanation:

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Can anyone help me out? I’ll give Brainly!! 5,175 isn’t correct :(

alexdok [17]

I’m sorry i can’t help :,( sending good grades for your last days of school!

3 0

3 years ago

Please show me how you also got the answer as well, thanks a bunch!!

egoroff_w [7]

Answer:

Step-by-step explanation:

4 0

2 years ago

What is the value of 2y + 3 when y +4 ?

MA_775_DIABLO [31]

Answer:

substitute y = 4 into the expression⇒(2×4)+3=8+3=11

Step-by-step explanation:

5 0

2 years ago

1 If a = p^1/3-p^-1/3<br>prove that: a^3 + 3a = p - 1/p

alexandr402 [8]

Hello, please consider the following.

We know that

$a = p^{\frac{1}{3}}-p^{-\frac{1}{3}}\\\\=p^{\frac{1}{3}}-\dfrac{1}{p^{\frac{1}{3}}}$

And we can write that.

$(p-\dfrac{1}{p})^3=(p-\dfrac{1}{p})(p^2-2+\dfrac{1}{p^2})\\\\=p^3-2p+\dfrac{1}{p}-p+\dfrac{2}{p}-\dfrac{1}{p^3}\\\\=p^3-\dfrac{1}{p^3}-3(p-\dfrac{1}{p})$

It means that, by replacing p by $p^{1/3}$

$(p^{1/3}-\dfrac{1}{p^{1/3}})^3=p-\dfrac{1}{p}-3(p^{1/3}-\dfrac{1}{p^{1/3}})\\\\\\\text{ So }\\\\a^3=p-\dfrac{1}{p}-3a\\\\\boxed{ a^3+3a=p-\dfrac{1}{p} }$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0

3 years ago

Solve for x. (PLEASE HELP)

nexus9112 [7]

9514 1404 393

Answer:

x = 6

Step-by-step explanation:

The parallel lines divide the sides proportionally. So, we can write the equation ...

x/4 = (x +3)/6

3x = 2(x +3) . . . . . multiply by 12

x = 6 . . . . . . . . . . . subtract 2x

7 0

3 years ago