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Maru [420]
3 years ago
9

You are given a problem that involves multiplying a positive number and a negative number.

Mathematics
2 answers:
posledela3 years ago
4 0

Answer: -3 and -2

Step-by-step explanation:

kenny6666 [7]3 years ago
3 0

Answer:

-3 and -2

Step-by-step explanation:

A better explanation...

(positive)(positive)= positive

(positive)(negative)= negative

(negative)(negative)= positive

(negative)(positive)= negative

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5 0
3 years ago
three terms of an arithmetic sequence are shown below. Which recursive formula defines the sequence? f(1) = 6, f(4) = 12, f(7) =
Viefleur [7K]
A = 6
tn = a + (n - 1)d
t4 = 6 + 3d = 12
3d = 12 - 6 = 6
d = 6/3 = 2
f(n + 1) = f(n) + 2
4 0
3 years ago
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PLZ HALP ME!!! eek<br><br> I will give brainliest, 5 stars, and a thanks
Elena L [17]

Answer: x = 6

Step-by-step explanation:

8 0
3 years ago
IF U HELP RN ILL GIVE U BRAINLIEST
shutvik [7]

Answer:

i need more info. that part dont help. if you took pic of the sentence they gv you that would be easier to work with.

Step-by-step explanation:

5 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
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