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3 years ago

Annesha is traveling at 50 mph. Morris is traveling 3 feet per second less than Aneesha. How fast is Morris traveling

Mathematics

1 answer:

inessss [21]3 years ago

4 0

Speed of Morris is 47.95 mph

Step-by-step explanation:

Step 1: Find the equivalent of Morris' speed in miles, that is convert 3 feet per second to miles per hour.

⇒ 3 ft/s = 3/1.467 mph (∵ 1 mph = 1.467 ft/s)

= 2.05 mph

Step 2: Find the speed of Morris.

Speed of Morris = 50 mph - 2.05 mph = 47.95 mph

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Solving Exponential Equations (lacking a common base)<br><br>(0.52)^q=4

bulgar [2K]

Answer:

q = log4 / log0.53

q = -2.189

Step-by-step explanation:

(0.53)^q = 4

Taking log of both sides!

q log 0.53 = log 4

q = log 4 / log 0.53

q = 0.602 / -0.275

q = - 2.189

This can be checked to confirm correctness.

Substituting q = -2.189

(0.53)^ -2.189 = 1/ 0.249

= 4.01 Proved!

(Note that the "1" is because of the negative sign)

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3 years ago

Read 2 more answers

When solving the system of equations by graphing, what is the solution of 3x + 2y = 2 and 2x – y=6?

oksian1 [2.3K]

Answer:

answer B: (2,-2)

Step-by-step explanation:

First, write the equations on top of each other:

$3x+2y=2\\2x-y=6$

Then, multiply the the second equation by 2 so that we can use elimination of the y-variable:

$3x+2y=2\\2(2x-y)=2(6)\\\\3x+2y=2\\4x-2y=12$

Next, use elimination to find the value of "x":

$3x+2y=2\\+(4x-2y=12)\\\\7x+0=14\\7x=14\\\frac{7x}{7}=\frac{14}{7}\\x=2$

So, your x-value is 2.

Now, substitute your x-value into one of your equations, let's take the second equation, 2x-y=6:

$2x-y=6\\2(2)-y=6\\4-y=6\\4-4-y=6-4\\-y=2\\\frac{-y}{1}=\frac{2}{-1}\\y=-2$

Your y-value is -2.

With all your information gathered, you find that the solution to this system of equation is (2,-2).

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3 years ago

Give the slope of each line. Also, state if the lines are parallel (yes or no).

VMariaS [17]

Answer:

AB slope is 1/2

CD slope is 1/3

Since the slopes of the lines aren't the same, the lines aren't parallel.

Step-by-step explanation:

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3 years ago

Explain what happens when you have an exponent to the power of another exponent.

jekas [21]

Answer:

The "power rule" tells us that to raise a power to a power, just multiply the exponents. Here you see that 52 raised to the 3rd power is equal to 56.

Step-by-step explanation:

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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

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

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3 years ago