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3 years ago

Which expression is equivalent to sine StartFraction 7 pi Over 6 EndFraction?

Mathematics

2 answers:

miskamm [114]3 years ago

7 0

Answer:

it is d the test sayed

Step-by-step explanation:

Nadya [2.5K]3 years ago

4 0

Answer:

Its d

Step-by-step explanation:

I plugged the stuff into photo math and once solved d was equal to it

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Please help!! Very urgent..thank you!!

erastova [34]

Answer : 4 times

Here it's given that ,

The height and base of the butterfly sitting on the stem (red butterfly) is two times greater than the height and base of the butterfly sitting on the flower .

And we need to find out how many times the area of red winged butterfly is greater than that of sitting on the flower (blue butterfly) .

Let us take ,

base of blue butterfly be b
height of blue butterfly be h
Area be A .

Then ,

base of red butterfly will be 2b .
height of red butterfly will be 2h .
Area be A' .

We know that ,

→ area of the triangle = 1/2 × base × height

So that ,

→ A/A' = (1/2 * b * h) ÷ (1/2 *2b *2h)

→ A/A' = bh/4bh

→ A/A' = 1/4

→ A' = 4A

Henceforth the area of blue butterfly is 4 times greater than that of red winged butterfly .

I hope this helps.

6 0

2 years ago

Rewrite the following equation in slope-intercept form.

KonstantinChe [14]

Answer:

y= 1/7x-6

Step-by-step explanation:

that is the answer in slope intercept form

8 0

2 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago

In a zoo there are 17 more monkeys than lions and 30 more lizards than lions. If there were 151 monkeys, lions and lizards in to

kvv77 [185]

If the amount of monkeys is m, lions is a, and lizards is b, then a+b+m=151, 17+a=m (since there are 17 more monkeys than lions), and 30+a=b. Substituting those values in (17+a=m and 30+a=b), we have 30+17+a+a+a=151=47+3a. Subtracting 47 from both sides, we get 104=3a. Next, we can divide both sides by 3 to get 104/3=a (the number of lions), 104/3+17=the number of monkeys, and 104/3+30=the number of lizards. Somehow we end up with a fraction (not a whole number), so there are 2/3rds of monkeys floating around using this logic.

8 0

3 years ago

Write the ordered pair that represents yz. Then find the magnitude of yz. Y(-2, 5), z(1,3)

Marysya12 [62]

Answer:

Okay, gimme a sec...

Step-by-step explanation:

8 0

3 years ago