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mixas84 [53]
3 years ago
7

4. Allison's vegetable garden is in shape of rectangle with an area of 10 feet^2. The width of

Mathematics
1 answer:
inna [77]3 years ago
6 0

Answer:

33333

Step-by-step explanation:

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Brandon has 32 stamps. He wants to display the stamps in rows, with the same number of stamps in each row. How many different wa
aniked [119]
Find all the factors of 32, as they come in pairs.
The pairs will give you how many rows you can have, as well as the number of stamps in each row.
1x32 would be 1 row, 32 stamps each
2x16 would be 2 rows, 16 stamps each.
4x8 would be 4 rows, 8 stamps each.
Once you reach the point that it flips around, you can count up all those ways as well.
8x4 is 8 rows with 4 stamps each
16x2 is 16 rows with 2 stamps each
32x1 is 32 rows with 1 stamp each
The total number of factor pairs is 6, meaning he can display the stamps in 6 different ways that follow his want of having the same number of stamps in each row.
8 0
3 years ago
Read 2 more answers
I need help in my homework here it is <br><br> Solve the equation for x.<br> x3 = 729
Andrej [43]

x3=729

729/3= 243

x=243

6 0
3 years ago
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GCF and LCM of Integers and Monomials
puteri [66]

Answer:

therdgsger

tgtrs

Step-by-step explanation:

3 0
3 years ago
1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.
xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}

(If you were to plot the actual curve, you would have both y=x^{2/3} and y=-x^{2/3}, but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx

We have

y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}

Then in the integral,

\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx

Substitute

u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx

This transforms the integral to

\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du

and computing it is trivial:

\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)

We can simplify this further to

\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}

7 0
3 years ago
Round 8,295.357 to the nearest tenth
Tresset [83]

Answer:

8,295.4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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