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3 years ago

Write the factored form of each Trinomial 31-35 please

Mathematics

2 answers:

Anarel [89]3 years ago

5 0

Answer:

not correct answer is it a question answer

alekssr [168]3 years ago

4 0

X^2-16xy+28y^2
I think this is correct

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What is the greatest common factor of 7 and 10

nasty-shy [4]

The answer is 1.
7 x 1 = 7
and 10 x 1 = 10
They don't have other factors in common besides 1.

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3 years ago

Read 2 more answers

A merchant sold a pen for $6.90 ,thereby making a profit of 15% on her cost calculate the cost of the pen to the merchant to the

NeTakaya

So you know the merchant made a 15% profit on the pen, so she bought it for a cheaper price. To find the cost of the pen before you have to take the price now, $6.90 and times it by 85%. You do 85% because you subtract the 15% she saved from 100% and you get 85%. So 6.90x.85= 5.865 which rounds to $5.87

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4 years ago

Mary wants to find this product. 58 x 20 How can Mary find the product? A. 0 +16 + 100 B. 0 + 160 + 1000 . C. 58 + 16 + 100 D. 5

Fittoniya [83]

Answer:

Step-by-step explanation:

58×20=1160

0+160+1000=1160

8 0

3 years ago

F(9) if f(x) = 5x - 16

MariettaO [177]

Here is your answer

6 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

4 years ago