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2 years ago

12

17/5 converted into a mixed number

2 answers:

meriva2 years ago

5 0

Answer:

3 2/5

Step-by-step explanation:

Mila [183]2 years ago

3 0

Answer:

3 $\frac{2}{5}$

Step-by-step explanation:

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The point P on the circle x2 + y2 = r2 that is also on the terminal side of an angle θ in standard position is given. Find the i

Brut [27]

Okay so you do not really need the circle equation. if you make a triangle with it's x length as 3 and it's y length as 4, you will be able to find the third length. Do Pythagorean theorem to find the hypotenuse.
the hypotenuse will be five after you calculate it.

sin is opposite over hypotenuse or Y over R

so... the sin is 4/5

make sure you know what quadrant your triangle is in for the negatives

6 0

3 years ago

John collected 17 leaves to feed his caterpillar collection if he wanted to split the leaves equally amongst the four cages how

Papessa [141]

Answer:

It lies between the numbers 4 and 5

Step-by-step explanation:

If John has 17 leaves and 4 cages then he should put 17/4, or 16/4+1/4 which is 4+0.25, or 4.25 in each cage.

3 0

3 years ago

Read 2 more answers

1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the

Alja [10]

I dont know all of them sorry. But i think
Its d. If its a list of questions lol. But I could be wrong but i did the math 2 times.

8 0

3 years ago

Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p

posledela

Answer:

<em>P=0.0000037</em>

<em>P=0.00037%</em>

Step-by-step explanation:

<u>Probability</u>

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

$\displaystyle \frac{4}{52}=\frac{1}{13}$

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

$\displaystyle \frac{3}{51}=\frac{1}{17}$

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

$\displaystyle \frac{2}{50}=\frac{1}{25}$

Finally, the probability of getting the remaining ace out of the 49 cards is:

$\displaystyle \frac{1}{49}$

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

$\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}$

$\displaystyle P= \frac{1}{270,725}$

P=0.0000037

P=0.00037%

3 0

3 years ago

Find the term indecent of x in the expansion of (x^2-1/x)^6

Mars2501 [29]

By the binomial theorem,

$\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}$

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0 ===> 3k = 12 ===> k = 4

which corresponds to the constant coefficient

$\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}$

3 0

3 years ago