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3 years ago

Using the following triangle, what is the cosine of angle B?

Mathematics

1 answer:

Nikolay [14]3 years ago

7 0

Answer:

cos(B) = a/c

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you that ...

Cos = Adjacent/Hypotenuse

The leg adjacent to angle B is "a". The hypotenuse is "c", so the desired cosine is ...

cos(B) = adjacent/hypotenuse = a/c

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SCORPION-xisa [38]

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3 years ago

Read 2 more answers

Use the number 3, 3, 5, and 6 only once, along with +, -, ×, ÷, or () to get a result of 3

gayaneshka [121]

Answer:

5 x 3 + 3 divided by 6 = 3

Step-by-step explanation:

The first thing is to find a way to arrange the numbers. Try every operation until you can get the answer. for this case, i put the answer above.

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2 years ago

Describe how to transform the quantity of the fifth root of x to the seventh power, to the third powerinto an expression with a

geniusboy [140]

$(\sqrt[5]{x^{7}})^{3}=(x^{\frac{7}{5}})^{3}=x^{\frac{7\cdot3}{5}}=x^{\frac{21}{5}}$

The root is equivalent to a fractional power with that number as the denominator. Otherwise, the rules of exponents apply.

7 0

3 years ago

Gimme some good memes, the best get Brainliest!

jeyben [28]

Roses are Red

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God made me bootiful

What the hell happed to you. 0,0

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This one sucks but :>

4 0

2 years ago

Read 2 more answers

A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

5 0

2 years ago