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3 years ago

Write the expression 8⋅ y ⋅ y ⋅ y ⋅ y using exponents.

Mathematics

2 answers:

Paladinen [302]3 years ago

6 0

Answer:

8^4 8 to the fourth power

Step-by-step explanation:

andrezito [222]3 years ago

4 0

Answer:

8 * y^4

Step-by-step explanation:

There are not like terms so you can't combine

y^4 = because there are 4 sets of y's

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The cost of renting a kayak for one hour is $23. Each additional hour is $8 more. Write the formulas to represent the situation.

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Step-by-step explanation:

Add and then multiply the one left

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3 years ago

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Nimfa-mama [501]

Answer:

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2 years ago

Simplify (6x2 + 11x − 3) + (2x2 − 17x − 4).

d1i1m1o1n [39]

Answer:

B. 8x^2-6x-7

Step-by-step explanation:

All you have to do is combine the like terms.

Like terms are the terms that have the same variable and same exponent.

The like terms in this equation are 6x^2 and 2x^2, 11x and -17x, and -3 and -4.

When you add 6x^2 and 2x^2, you get 8x^2

When you add 11x and -17x, you get -6x

When you add -3 and -4, you get -7.

Putting these all in order, your answer is

8x^2 - 6x - 7

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2 years ago

Two sides of a triangle measure 8cm and 12cm which can not be the measure of the third side?

seraphim [82]

4cm, as if two of the sides add up to be the same length as the first side then the triangle would be a straight line.

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3 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