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3 years ago

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

1 answer:

8 0

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

Step-by-step explanation:

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:

find the length of each side:

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

now find the area of a triangle:

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: $\frac{3\sqrt{3} }{2} *a^2$

7 0

3 years ago

Read 2 more answers

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.

6 0

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

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

5 0

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=

7 0

3 years ago