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3 years ago

Given cos (theta)=1, find cot (theta)

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

6 0

First problem:

cos (theta)=1

Using the inverse cosine function, you get theta = 0.

Now we find tan 0 = 0

cot(theta) = 1/tan(theta) = 1/0

Division by zero is undefined, so the answer is d. undefined

Second problem:

cos (theta)=1

Use the inverse cosine function.

theta = 0°

Answer: c. 0°

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This is another one helpzzz mee

Varvara68 [4.7K]

I don't know about the first one but the second one is 34.9°. It would be that because are right angles add up to 90° and if you subract 55.1 and 90 it would equal 34.9. But this rule only works on right angles.

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3 years ago

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6x + 4y=10 12x + y= -29

Olin [163]

Answer: x=-3 y=7.

Step-by-step explanation:

$\displaystyle\\\left \{ {{6x+4y=10\ \ \ \ \ (1)} \atop {12x+y=-29\ \ \ \ (2)}} \right.\\ We\ multiply\ both\ sides\ of\ equation\ (1)\ by\ -2:\\\left \{ {{(-2)*6x+(-2)*4y=(-2)*10} \atop {12x+y=-29}} \right. \\\\\left \{ {{-12x-8y=-20} \atop {12x+y=-29}} \right. \\$ $We\ summarize\ equations\ (1)\ and\ (2):\\-7y=-49\\-7*y=-7*7\\Divide\ both\ sides\ of\ the\ equation\ by\ -7:\\y=7.\\We\ substitute\ the\ value\ of\ y\ into \ equation\ (1):\\6x+4*7=10\\6x+28=10\\6x+28-28=10-28\\6x=-18\\6*x=6*(-3)\\Divide\ both\ sides\ of\ the\ equation\ by\ 6:\\x=-3.$

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2 years ago

5(-1x+5)=10 pls help

Leya [2.2K]

Answer:

-5x+5=10

-5x=10-5

-5x=5

x=-5/5

x=-1

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3 years ago

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Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5