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Dahasolnce [82]
3 years ago
9

The quotient of the sum of 2t and 2 and twice the cube of s

Mathematics
1 answer:
Vladimir79 [104]3 years ago
8 0

The equation according to the given verbal sentence is \frac{2 t+2}{2 s^{3}} (after solving this \frac{t+1}{s^{3}} ).

<u>Step-by-step explanation:</u>

Rewrite the given verbal sentence and so it will be easier to translate.

Given: The quotient of the sum of 2 t and 2 and twice the cube of s

sum of 2 t and 2 can be written as

       2 t + 2

twice the cube of s

      2 s^{3}

Now, combine the above according to the given verbal. The quotient of the sum of 2 t and 2 and twice the cube of s  means,

             2 t + 2  divided by 2 s^{3}

Now, the equation would be \frac{2 t+2}{2 s^{3}}

When solving this equation, we get

         \frac{2(t+1)}{2 s^{3}}=\frac{t+1}{s^{3}}

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Help! This is due today and I'm stuck on this problem.
Varvara68 [4.7K]

Answer:

280.6 m^2

Step-by-step explanation:

<u>find the length of each side:</u>

you're only given h = 9 m

the formula for h is  h=\frac{\sqrt{3} }{2} *a, where a is a side length of the hexagon

plug h in:

9=\frac{\sqrt{3}}{2}*a\\\\a=9*\frac{2}{\sqrt{3}}\\\\

a ≈ 10.392 m

<u>now find the area of a triangle:</u>

1/2 (10.392)(9) = 46.764 m^2

multiply that by 6 and you get 280.584 ≈ 280.6 m^2

you can also use the regular hexagon area formula instead of finding the individual area and multiplying by 6: \frac{3\sqrt{3} }{2} *a^2

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3 years ago
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When a figure is translated, what is true, compared to the preimage of the figure
Charra [1.4K]
When you translate something in geometry, you're simply moving it around. You don't distort it in any way. If you translate a segment, it remains a segment, and its length doesn't change. Similarly, if you translate an angle, the measure of the angle doesn't change.
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2 years ago
Please ASAP i need help
NeX [460]

Answer:

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Step-by-step explanation:

easier to understand

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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)
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3 years ago
Need the answer ASAP
Evgesh-ka [11]
Anwser:

22.5 or 5/22 or 30

Step-by-step Explanation:

-2 and -2
SSS
-2+-2= -4

-4 x 30= -120

-120 divided by 4 is 30
but that’s not a fraction so you do

22.5 or 5/22

-2 x -2 x 30=

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