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3 years ago

Select the correct answer.

Mathematics

2 answers:

slamgirl [31]3 years ago

8 0

Answer:

balls

Step-by-step explanation:

balls 2

babymother [125]3 years ago

8 0

Answer:

haha

Step-by-step explanation:

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Morgan rolls a number cube, twice. If the number 1 shows up at least once, Morgan wins. Otherwise, Jaclyn wins. How much greater

alukav5142 [94]

I am assuming from the answer choices that you meant Jaclyn's chances of winning.

There are 6 sides of a dice and it's rolled twice.
6×2=12

So while Morgan has a 2/12 chance (simplified that would be 1/6), Jaclyn has a 10/12 chance of winning (simplified that would be 1/3).

Therefore your answer would be A: 1/3

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3 years ago

11.5 is what percent of 8? Round to the nearest tenth.

Alik [6]

Answer:

143.8% is the answer

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2 years ago

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Green Leaf Lawn Care had 64 customers under contract at the start of the year. The company's owner expects his new radio adverti

cestrela7 [59]

Answer:

C. c = 64 + 2w

Step-by-step explanation:

You're starting with 64 customers and you're gaining 2 for each week.

good luck, i hope this helps :)

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2 years ago

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Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$