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3 years ago

Okay help again u guys are so smart

Mathematics

1 answer:

ra1l [238]3 years ago

3 0

Answer:

1 year ago question im sorry free point

Step-by-step explanation:

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If one card is drawn from a standard deck of 52 playing cards what is the probability of it is not a heart

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I forgot how many of each card there are and if you're counting the joker.
But just find the number of heart cards, and minus that from 52.
The answer divided by 52 is the probability that it's not a heart.

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3 years ago

What is another way to group the factors (3x2)x5

tia_tia [17]

<u>Answer</u>

3×(2×5)

<u>Explanation</u>

Multiplication of numbers is associative. For example,

(a×b)×c = a×(b×c)

This is also called grouping. We multiply more than 2 numbers by grouping.

For the equation given above, (3x2)x5, it can also be grouped as 3×(2×5).

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What is 19 divided by 3192

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The answer is 168!!!!!!!!!!!!!!!!!!!

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Find the exact value of the expression.

Tresset [83]

$\sin(a-b)=\sin a \cos b-\cos a \sin b$ . If we let $a=\cos^{-1} \left(\frac{5}{6} \right)$ and $b=\tan^{-1} \left(\frac{1}{2} \right)$ , then the given expression is equal to:

$\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)$

Using the Pythagorean identities $\sin^{2} x+\cos^{2} x=1$ and $\tan^{2} x+1=\sec^{2} x$ ,

$1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}$

$2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

$\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

This means we can write the original expression as:

$\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}$

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