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3 years ago

6

The price of one type of cheese is $2.60 per pound. What is the cost of a 2.5-pound package of cheese?

1 answer:

Yuki888 [10]3 years ago

6 0

The price would be 2.60*2.5=$6.50

Just multiply the price cause it’s by the pound and multiply it by how many pounds

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First change to standard form of first order linear differential equation and then by finding the appropriate integrating factor

Leokris [45]

Answer:

Step-by-step explanation:

Given is a Differential equation as

$xy' + y = lnx ; y(e)=1$

To bring it to linear form we can divide the full equation by x

$y'+\frac{y}{x} =\frac{ln x}{x}$

This is of the form

y'+p(x) *y = q(x)

p(x) = 1/x

So find

$e^{\int\limits{\frac{1}{x} } \, dx } = e^{ln x} = x$

Solution is

$xy = \int {x*lnx /x } \, dx =xln x -x +C$

Use the initial value as y(e) =1

$e= eln e -e+C\\C=e$

So solution is

$xy =xln x -x+e$

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3 years ago

This question please; I’ve done all the others but this one has stumped me :/

vesna_86 [32]

2 way table
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boy 12 4 16 32
girl 14 9 5 28
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3 years ago

What is the square root of 3 to the square root of 5?

Tresset [83]

Square root of three- 1.73205
Square root of five- 2.23607

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3 years ago

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What is the value of n when (9)^2n-1 = (27)^n+2

Elis [28]

$9^{2n-1} = 27^{n+2} \\ ( {3}^{2} ) ^{2n - 1} = ( {3}^{3} )^{ n + 2} \\ {3}^{4n - 2} = {3}^{3n + 6} \\ 4n - 2 = 3n + 6 \\ 4n - 3n = 6 + 2 \\ n = 8$

Answer: n=8

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3 years ago

The portion of the parabola y²=4ax above the x-axis, where is form 0 to h is revolved about the x-axis. Show that the surface ar

castortr0y [4]

Answer:

See below for Part A.

Part B)

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

Step-by-step explanation:

Part A)

The parabola given by the equation:

$y^2=4ax$

From 0 to <em>h</em> is revolved about the x-axis.

We can take the principal square root of both sides to acquire our function:

$y=f(x)=\sqrt{4ax}$

Please refer to the attachment below for the sketch.

The area of a surface of revolution is given by:

$\displaystyle S=2\pi\int_{a}^{b}r(x)\sqrt{1+\big[f^\prime(x)]^2} \,dx$

Where <em>r(x)</em> is the distance between <em>f</em> and the axis of revolution.

From the sketch, we can see that the distance between <em>f</em> and the AoR is simply our equation <em>y</em>. Hence:

$r(x)=y(x)=\sqrt{4ax}$

Now, we will need to find f’(x). We know that:

$f(x)=\sqrt{4ax}$

Then by the chain rule, f’(x) is:

$\displaystyle f^\prime(x)=\frac{1}{2\sqrt{4ax}}\cdot4a=\frac{2a}{\sqrt{4ax}}$

For our limits of integration, we are going from 0 to <em>h</em>.

Hence, our integral becomes:

$\displaystyle S=2\pi\int_{0}^{h}(\sqrt{4ax})\sqrt{1+\Big(\frac{2a}{\sqrt{4ax}}\Big)^2}\, dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax}\Big(\sqrt{1+\frac{4a^2}{4ax}}\Big)\,dx$

Combine roots;

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax\Big(1+\frac{4a^2}{4ax}\Big)}\,dx$

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax+4a^2}\, dx$

Integrate. We can consider using u-substitution. We will let:

$u=4ax+4a^2\text{ then } du=4a\, dx$

We also need to change our limits of integration. So:

$u=4a(0)+4a^2=4a^2\text{ and } \\ u=4a(h)+4a^2=4ah+4a^2$

Hence, our new integral is:

$\displaystyle S=2\pi\int_{4a^2}^{4ah+4a^2}\sqrt{u}\, \Big(\frac{1}{4a}\Big)du$

Simplify and integrate:

$\displaystyle S=\frac{\pi}{2a}\Big[\,\frac{2}{3}u^{\frac{3}{2}}\Big|^{4ah+4a^2}_{4a^2}\Big]$

Simplify:

$\displaystyle S=\frac{\pi}{3a}\Big[\, u^\frac{3}{2}\Big|^{4ah+4a^2}_{4a^2}\Big]$

FTC:

$\displaystyle S=\frac{\pi}{3a}\Big[(4ah+4a^2)^\frac{3}{2}-(4a^2)^\frac{3}{2}\Big]$

Simplify each term. For the first term, we have:

$\displaystyle (4ah+4a^2)^\frac{3}{2}$

We can factor out the 4a:

$\displaystyle =(4a)^\frac{3}{2}(h+a)^\frac{3}{2}$

Simplify:

$\displaystyle =8a^\frac{3}{2}(h+a)^\frac{3}{2}$

For the second term, we have:

$\displaystyle (4a^2)^\frac{3}{2}$

Simplify:

$\displaystyle =(2a)^3$

Hence:

$\displaystyle =8a^3$

Thus, our equation becomes:

$\displaystyle S=\frac{\pi}{3a}\Big[8a^\frac{3}{2}(h+a)^\frac{3}{2}-8a^3\Big]$

We can factor out an 8a^(3/2). Hence:

$\displaystyle S=\frac{\pi}{3a}(8a^\frac{3}{2})\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Simplify:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

Hence, we have verified the surface area generated by the function.

Part B)

We have:

$y^2=36x$

We can rewrite this as:

$y^2=4(9)x$

Hence, a=9.

The surface area is 1000. So, S=1000.

Therefore, with our equation:

$\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]$

We can write:

$\displaystyle 1000=\frac{8\pi}{3}\sqrt{9}\Big[(h+9)^\frac{3}{2}-9^\frac{3}{2}\Big]$

Solve for h. Simplify:

$\displaystyle 1000=8\pi\Big[(h+9)^\frac{3}{2}-27\Big]$

Divide both sides by 8π:

$\displaystyle \frac{125}{\pi}=(h+9)^\frac{3}{2}-27$

Isolate term:

$\displaystyle \frac{125}{\pi}+27=(h+9)^\frac{3}{2}$

Raise both sides to 2/3:

$\displaystyle \Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}=h+9$

Hence, the value of h is:

$\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614$

8 0

3 years ago

Read 2 more answers