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Nimfa-mama [501]
3 years ago
12

When a number is raised to the 6th power, it equals 90. What is this number, and write it using the root symbol ...?

Mathematics
1 answer:
balandron [24]3 years ago
6 0
6√90 = <span>2.11693286302546
So, this number to the power of six equals 90. 
</span><span>2.11693286302546^6 = 90
</span>
The correct answer is <span>2.11693286302546. </span>
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Solve sinX+1=cos2x for interval of more or equal to 0 and less than 2pi
Igoryamba

Answer:

Question 1: \sin(x)+1=\cos^2(x)

Answer to Question 1: x=0, \pi \frac{3\pi}{2}

Question 2: \sin(x)+1=\cos(2x)

Answer to Question 2: 0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}

Question:

I will answer the following two questions.

Condition: 0\le x

Question 1: \sin(x)+1=\cos^2(x)

Question 2: \sin(x)+1=\cos(2x)

Step-by-step explanation:

Question 1: \sin(x)+1=\cos^2(x)

Question 2: \sin(x)+1=\cos(2x)

Question 1:

\sin(x)+1=\cos^2(x)

I will use a Pythagorean Identity so that the equation is in terms of just one trig function, \sin(x).

Recall \sin^2(x)+\cos^2(x)=1.

This implies that \cos^2(x)=1-\sin^2(x). To get this equation from the one above I just subtracted \sin^2(x) on both sides.

So the equation we are starting with is:

\sin(x)+1=\cos^2(x)

I'm going to rewrite this with the Pythagorean Identity I just mentioned above:

\sin(x)+1=1-\sin^2(x)

This looks like a quadratic equation in terms of the variable: \sin(x).

I'm going to get everything to one side so one side is 0.

Subtracting 1 on both sides gives:

\sin(x)+1-1=1-\sin^2(x)-1

\sin(x)+0=1-1-\sin^2(x)

\sin(x)=0-\sin^2(x)

\sin(x)=-\sin^2(x)

Add \sin^2(x) on both sides:

\sin(x)+\sin^2(x)=-\sin^2(x)+\sin^2(x)

\sin(x)+\sin^2(x)=0

Now the left hand side contains terms that have a common factor of \sin(x) so I'm going to factor that out giving me:

\sin(x)[1+\sin(x)]=0

Now this equations implies the following:

\sin(x)=0 or 1+\sin(x)=0

\sin(x)=0 when the y-coordinate on the unit circle is 0. This happens at 0, \pi, or also at 2\pi. We do not want to include 2\pi because of the given restriction 0\le x.

We must also solve 1+\sin(x)=0.

Subtract 1 on both sides:

\sin(x)=-1

We are looking for when the y-coordinate is -1.

This happens at \frac{3\pi}{2} on the unit circle.

So the solutions to question 1 are 0,\pi,\frac{3\pi}{2}.

Question 2:

\sin(x)+1=\cos(2x)

So the objective at the beginning is pretty much the same. We want the same trig function.

\cos(2x)=\cos^2(x)-\sin^2(x) by double able identity for cosine.

\cos(2x)=(1-\sin^2(x))-\sin^2(x) by Pythagorean Identity.

\cos(2x)=1-2\sin^2(x) (simplifying the previous equation).

So let's again write in terms of the variable \sin(x).

\sin(x)+1=\cos(2x)

\sin(x)+1=1-2\sin^2(x)

Subtract 1 on both sides:

\sin(x)+1-1=1-2\sin^2(x)-1

\sin(x)+0=1-1-2\sin^2(x)

\sin(x)=0-2\sin^2(x)

\sin(x)=-2\sin^2(x)

Add 2\sin^2(x) on both sides:

\sin(x)+2\sin^2(x)=-2\sin^2(x)+2\sin^2(x)

\sin(x)+2\sin^2(x)=0

Now on the left hand side there are two terms with a common factor of \sin(x) so let's factor that out:

\sin(x)[1+2\sin(x)]=0

This implies \sin(x)=0 or 1+2\sin(x)=0.

The first equation was already solved in question 1. It was just at x=0.

Let's look at the other equation: 1+2\sin(x)=0.

Subtract 1 on both sides:

2\sin(x)=-1

Divide both sides by 2:

\sin(x)=\frac{-1}{2}

We are looking for when the y-coordinate on the unit circle is \frac{-1}{2}.

This happens at \frac{7\pi}{6} or also at \frac{11\pi}{6}.

So the solutions for this question 2 is 0,\pi,\frac{7\pi}{6}, \frac{11\pi}{6}.

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In how many different ways can a ski club consisting of 20 people select a person for its officers? the positions available are
LUCKY_DIMON [66]
There are 116,280 ways to select the officers.

Since order is important, we use permutations:

P(20,4)=\frac{20!}{(20-4)!}=\frac{20!}{16!}=116280
3 0
3 years ago
Charlotte bought 25 cookies to share with her class, and 5 of them are oatmeal raisin. What percent of the cookies are oatmeal r
LUCKY_DIMON [66]

Answer:

20%

Step-by-step explanation:

25 divided by 5 = 5

Convert to decimal

0.20

Convert to percentage

20%

7 0
3 years ago
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It is about solving equations by substitution. if you can help plzzz
Lyrx [107]
Message me, and i can help more:)
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How can you solve this problem <br> 2250=5000 x R x 5
vesna_86 [32]
R is equal to 450

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