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3241004551 [841]
3 years ago
6

4. Compare using <> or = -5 O -9

Mathematics
1 answer:
tatiyna3 years ago
7 0

Answer:

-9 <-5 <0

Step-by-step explanation:

hope it helps

for any question comment me❤❤

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Tony has 7/8 of a candy bar. he wants to split the bar into servings that are 3/10 of the bar. how many servings can he make?
erica [24]
Alright, so let's make both 7/8 and 3/10 into fractions with the same denominator.
The lowest common denominator for both of these is 80.
So, we multiply the top and bottom of 7/8 by 10, and the top and bottom of 3/10 by 8 to get fractions we can compare.
So, we have 70/80 of a candy bar, and we want to make servings of 24/80 each.
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3 years ago
GEOMETRY: You are building a shelf that fits in a corner. In the figure, the entire shelf is IJK . Each unit in the coordinate p
arsen [322]

Answer:

A= 12.5 inches²

Step-by-step explanation:

a=1/2bh

a=1/2(5)5

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a= 12.5

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
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