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3 years ago

Hi do any of you know how to do this?

Mathematics

2 answers:

Oksi-84 [34.3K]3 years ago

7 0

Answer: yes, it's a quadratic equation, the formula is x = −b ± √(b2 − 4ac) 2a. a=8m2 b=-2m and c =0. This only works if the other side is 0. The answer would be m=1!!!!! can i plz get a brainliest!

Step-by-step explanation:

Kamila [148]3 years ago

5 0

Answer: i got m=1

Step-by-step explanation:

8(1)^2-2(1)=8-2=6

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4vir4ik [10]

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8 0

3 years ago

Half the sum of seven and w?

swat32

Answer:

3.5w

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Order these decimals from smallest to largest 1.89 -0.81 -2.6 0.24 -1.7

tiny-mole [99]

Answer:

-2.6, -1.7, -0.81, 0.24, 1.89

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Reserve Problems Chapter 4 Section 5 Problem 2 A laptop company claims up to 10.1 hours of wireless web usage for its newest lap

Viktor [21]

Answer:

0.1151

Step-by-step explanation:

Given that a laptop company claims up to 10.1 hours of wireless web usage for its newest laptop battery life. However, reviews on this laptop shows many complaints about low battery life. A survey on battery life reported by customers shows that it follows a normal distribution with mean 9.5 hours and standard deviation 30 minutes.

If X is the battery life then X is N(9.5, 0.5)

(we convert into one unit for calculations purpose here hours)

a) the probability that the battery life is at least 10.1 hours

= $P(X\geq 10.1)\\=P(Z\geq \frac{10.1-9.5}{0.5} )\\= P(Z\geq 1.2)\\=0.1151$

we get the probability that the battery life is at least 10.1 hours is 0.1151

4 0

3 years ago