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3 years ago

In the diagram, angle OLM is twice as large as angle PON.

Mathematics

1 answer:

MrRa [10]3 years ago

4 0

Answer:

umm i not sure, but this is a guess because i am stuck on it too

Step-by-step explanation:

i think it is 106

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If janie has 3045 apples and Lilly had 4599 apples what's the difference

marin [14]

Answer: 1554 apples

Step-by-step explanation: In order to find the distance you have to subtract the biggest number from the smallest number.

4599 - 3045 = 1554 apples

8 0

3 years ago

PLS HELP THIS IS A TEST

Pie

Answer:

D: 7

Step-by-step explanation:

1.75 (7) + 4.50

12.25 + 4.50

= 16.75

16.75 < 20

8 0

3 years ago

For the data in the table, does y vary directly with x? If it does, write an equation for the direct variation.

Trava [24]

11 = 1.375 * 8
22 = 1.375 * 16
33 = 1.375 * 24
Therefore y varies directly with x.
Answer: C ) yes; y = 1.375 x

7 0

4 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 y is a function of x :

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

(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 ≤ x ≤ 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:

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