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3 years ago

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A record album is 3/16 of an inch thick. How many albums can be stacked to fit in a box 12 inches high? Please show me how you g

ot you answer explain

1 answer:

german3 years ago

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To the nearest hundredth, what is the value of x? <br> 36.08<br> 41.51<br> 47.81<br> 72.88

Crazy boy [7]

Answer:

B

Step-by-step explanation:

Using the law of sines, we can make a proportion.

But first, we'll need to solve for the unknown angle.

We add up the two known angles and subtract that by 180.

90 + 41 = 131

180 - 131 = 49

So the unknown angles is 49.

Then, we can use the law of sines.

Make the equation.

sin(90)/55 = sin(49)/x

Simplify this using a calculator and you get around 41.51 or option B.

5 0

3 years ago

What is the y-intercept of y = -2x+6y=−2x+6?

goblinko [34]

Answer:

$6$

Step-by-step explanation:

Given: $y=-2x+6$

Since the equation is written in slope-intercept form, it is extremely easy to find the y-intercept. Slope-intercept form is written as:

$y=mx+b$

Whereas:

y: range

m: slope

x: domain

b: y-intercept

With that in mind, we can immediately tell that 6 is the y-intercept.

6 0

3 years ago

Read 2 more answers

How many unit cubes would you need to make a model of a rectangular prism that is 4 units long × 3 units wide × 2 units high?

lara [203]

Answer:

C

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Two cube-shaped blocks are placed on a table. Both cubes have a mass of 3 kilograms. Cube 1 has a bottom surface area of 0.9 m²,

k0ka [10]

Answer:

The cube with surface area of 0.3 sq. meters will exert more pressure.

Step-by-step explanation:

Pressure is given by $P=\frac{mass*acceleration}{area}$

Since mass(m) and acceleration (a) for both the cubes are equal and the area varies, we will have to base our answer on the area.

In the formula for pressure, we can see that Pressure and Area are inverse of each other <em><u>(given mass * acceleration is constant, in this case).</u></em>

So the cube that will exert more pressure is the one with less surface area <u><em>(pressure and area are inverse of each other).</em></u>

Hence the cube with surface area of 0.3 sq. meters will exert more pressure.

8 0

4 years ago

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago