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3 years ago

Rewrite with only sin x and cos x. sin 3x

1 answer:

8 0

$\bf sin({{ \alpha}} + {{ \beta}})=sin({{ \alpha}})cos({{ \beta}}) + cos({{ \alpha}})sin({{ \beta}})
\\\\\\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\
cos(2\theta)=
\begin{cases}
\boxed{cos^2(\theta)-sin^2(\theta)}\\
1-2sin^2(\theta)\\
2cos^2(\theta)-1
\end{cases}\\\\
-------------------------------\\\\
sin(3x)\implies sin(2x+x)\implies sin(2x)cos(x)+cos(2x)sin(x)$

$\bf 2sin(x)cos(x)cos(x)~+~[cos^2(x)-sin^2(x)]sin(x)
\\\\\\
\stackrel{like~terms}{\stackrel{\downarrow }{2sin(x)cos^2(x)}~~+~~\stackrel{\downarrow }{sin(x)cos^2(x)}}-sin^3(x)
\\\\\\
3sin(x)cos^2(x)-sin^3(x)$

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Pls help me!! I just need help on both of the following questions, a-d on each

Ymorist [56]

I'll do a and e on the first one to show you how to do it

Then See if you can do the rest

a. 6(-3)

When your multiplying or dividing with negative numbers; If one of your numbers is negative like the one above, your answer will be negative

So 6 × 3 = 18 ; but since there is a negative on the 3 your answer will be -18

e. -20 ÷ 5

same for this; 20 ÷ 5 = 4, but since the 20 is negative the answer will be -4

You do b, c, d, and f on the first question. Tell me what you get and we'll go from there

Now for the second question 

I will do a and d to show you how to do this then you do the rest (I'm inserting a picture with these)

a. $4(-2 \frac{1}{3}) + 7 \frac{1}{6}$
First go ahead and turn the fractions into improper fractions, this makes them a bit easier to work with. When multiplying fractions; they do not need a common denominator, all you do is multiply top by top and bottom by bottom.
However, when you add and subtract fractions you MUST have a common denominator

d. (-4.25)(2) + (-4.25)(-2)
remember from the first problem a negative times a positive gives you a negative

but when you times a negative by a negative you get a positive

Now you try b and c; let me know what you get, and we'll go from there!

7 0

3 years ago

How much can this container hold?

nydimaria [60]

Answer:

225 in.^3

Step-by-step explanation:

You need to find the volume of the container.

The container has the shape of a rectangular prism.

volume = length * width * height

volume = 5 in. * 4.5 in. * 10 in.

volume = 225 in.^3

6 0

3 years ago

g Suppose that a die is rolled twice. What are the possible values that thefollowing random variables can take on:(a) the maximu

KengaRu [80]

Answer:

(a) A = {1, 2, 3, 4, 5, 6}

(b) B = {1, 2, 3, 4, 5, 6}

(d) D = {-5, -4, -3, -2, -1, -0, 1, 2, 3, 4, 5}

Step-by-step explanation:

Assume each roll can result in the numbers 1, 2, 3, 4, 5, or 6.

(a) If both rolls result in a 1, the maximum value is 1. If either roll results in a 6, the maximum value is 6; all integers between 1 and 6 are also possible. Therefore, the possible values are:

A = {1, 2, 3, 4, 5, 6}

(b) If either roll results in a 1, the minimum value is 1. If both rolls result in a 6, the minimum value is 6; all integers between 1 and 6 are also possible. Therefore, the possible values are:

B = {1, 2, 3, 4, 5, 6}

(c) If both rolls result in a 1, the sum is 2. If both rolls results in a 6, the sum is 12; all integers between 2 and 12 are also possible. Therefore, the possible values are:

C = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(d) If the first roll results in a 1 and the second results in a 6, the result is -5. On the other hand, if the first roll results in a 6 and the second results in a 1, the result is 5; all integers between -5 and 5 are also possible. Therefore, the possible values are:

D = {-5, -4, -3, -2, -1, -0, 1, 2, 3, 4, 5}

7 0

2 years ago

Proportions happen because two events or scenarios are ___

Anni [7]

Proportions happen because two events or scenarios are related or associated with one another

6 0

3 years ago

An alarming number of U.S. adults are either overweight or obese. The distinction between overweight and obese is made on the ba

madreJ [45]

Answer:

(A) The probability that a randomly selected adult is either overweight or obese is 0.688.

(B) The probability that a randomly selected adult is neither overweight nor obese is 0.312.

(D) The events "overweight" and "obese" mutually exclusive.

Step-by-step explanation:

Denote the events as follows:

X = a person is overweight

Y = a person is obese.

The information provided is:

A person is overweight if they have BMI 25 or more but below 30.

A person is obese if they have BMI 30 or more.

P (X) = 0.331

P (Y) = 0.357

(A)

The events of a person being overweight or obese cannot occur together.

Since if a person is overweight they have (25 ≤ BMI < 30) and if they are obese they have BMI ≥ 30.

So, P (X ∩ Y) = 0.

Compute the probability that a randomly selected adult is either overweight or obese as follows:

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\=0.331+0.357-0\\=0.688$

Thus, the probability that a randomly selected adult is either overweight or obese is 0.688.

(B)

Commute the probability that a randomly selected adult is neither overweight nor obese as follows:

$P(X^{c}\cup Y^{c})=1-P(X\cup Y)\\=1-0.688\\=0.312$

Thus, the probability that a randomly selected adult is neither overweight nor obese is 0.312.

(C)

If two events cannot occur together, but they form a sample space when combined are known as exhaustive events.

For example, flip of coin. On a flip of a coin, the flip turns as either Heads or Tails but never both. But together the event of getting a Heads and Tails form a sample space of a single flip of a coin.

In this case also, together the event of a person being overweight or obese forms a sample space of people who are heavier in general.

Thus, the events "overweight" and "obese" exhaustive.

(D)

Mutually exclusive events are those events that cannot occur at the same time.

The events of a person being overweight and obese are mutually exclusive.

5 0

2 years ago