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4 years ago

What is the cost per square foot for a (50 ft. x 230 ft.) rectangular lot that is selling for $8625.

1 answer:

6 0

First you have to find the area, which is 11,500ft. Then you divide the cost of the land by the area (8625/11500) which equals 0.75. So it is 75 cents per square foot of land.
Hope this helped!

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3(x-1)=2x+9 how many solutions are there

Nostrana [21]

There are 4 solutions

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3 years ago

The area of a rectangular field is 8811 m2. if the width of the field is 89m, what is the length?

xeze [42]

Length = 99 because...
since you multiply to get the area, in order to find the missing term of an area (in this case its length), you do the opposite of what you started with...so if you multiplied to get the area, you need to divide to get the missing side...
so 8811 divided by 89 = 99
99 x 89 = 8811

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4 years ago

The base of a cube is 1/4 ft squared. Whats the volume of the cube? also can you explain your answer so i can answer other quest

MrMuchimi

It would be 1/64 I hope this helps with your question.

8 0

3 years ago

Read 2 more answers

How to use the slope formula to find the slope of a line<br> please help me!!!!!!!

stepan [7]

Answer:

Sweetie, it's easy. Let me explain.

The slope formula is (y2 - y1) (x2 - x1) you find the slope by putting the rise over the run. For example, if the point goes up 2 and right 3 your slope would be (2,3) and let's just say your next coordinate is (4,5) you would take the y, 5, from the second coordinate and subtract it by the y from the first coordinate, 3. So you would have 2 and you do the same with the x values. So your slope would be 3/2. Hope I helped at all.

Step-by-step explanation:

8 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

4 years ago