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AlexFokin [52]
3 years ago
5

What is the value of 4530 divided by 10³ A. 0.453 B. 4.530 C. 45.30 D. 453.0 

Mathematics
2 answers:
Ilya [14]3 years ago
6 0

We first would have to figure out what would 10 to the power of 3 is.

10 x 10 x 10 = 1,000

Then, we do what the problem is really asking.

4,530 ÷ 1,000 = 4.53

Your answer: B. 4.530

goblinko [34]3 years ago
4 0
The value of 4530 divided by 10^3 is B
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There are 15 people on a basketball team, and the coach needs to choose 5 to put into
snow_lady [41]

Answer:

The coach can do this in 3,003 ways

Step-by-step explanation:

Here, the coach needs to select a team of 5 from a total of 15 players

Mathematically, the number of ways this can be done is simply 15 C5 ways

Generally, if we are to select a number of r items from n items, this can be done in nCr ways = n!/(n-r)!r!

Applying this to the situation on ground, we have;

15C5 = 15!/(15-5)!5! = 15!/10!5! = 3,003 ways

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3 years ago
The common ratio in a geometric series is 0.50.50, point, 5 and the first term is 256256256.
V125BC [204]

Answer:

504

Step-by-step explanation:

I think the correct question is like:The common ratio in a geometric series is 0.50 ( point 5) and the first term is 256.  

Find the sum of the first 6 terms in the series.

If it's right, then

Sₙ = (a₁ * (1 - rⁿ)) / (1 - r)

S₆ = (256 * ( 1 - 0.5⁶)) / (1 - 0.5) = (256 * 0.984375) / 0.5 = 504

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3 years ago
What is true about the interior and exterior angle measures of any triangle
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Interior angles of a triangle=180
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Ingrid walked her dog and washed her car. The time she spent walking her dog was one-fourth the time it took her to wash her car
STALIN [3.7K]
1/4 of 14 is 7/2 which is 3 1/2 which is 3.5
4 0
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Read 2 more answers
Is anybody else here to help me ??​
Akimi4 [234]

Answer:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

Step-by-step explanation:

I'm going to use x instead of \theta because it is less characters for me to type.

I'm going to start with the left hand side and see if I can turn it into the right hand side.

\cot(x)+\cot(\frac{\pi}{2}-x)

I'm going to use a cofunction identity for the 2nd term.

This is the identity: \tan(x)=\cot(\frac{\pi}{2}-x) I'm going to use there.

\cot(x)+\tan(x)

I'm going to rewrite this in terms of \sin(x) and \cos(x) because I prefer to work in those terms. My objective here is to some how write this sum as a product.

I'm going to first use these quotient identities: \frac{\cos(x)}{\sin(x)}=\cot(x) and \frac{\sin(x)}{\cos(x)}=\tan(x)

So we have:

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

I'm going to factor out \frac{1}{\sin(x)} because if I do that I will have the \csc(x) factor I see on the right by the reciprocal identity:

\csc(x)=\frac{1}{\sin(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

Now I need to somehow show right right factor of this is equal to the right factor of the right hand side.

That is, I need to show \cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)} is equal to \csc(\frac{\pi}{2}-x).

So since I want one term I'm going to write as a single fraction first:

\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)}

Find a common denominator which is \cos(x):

\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}

\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}

\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}

By  the Pythagorean Identity \cos^2(x)+\sin^2(x)=1 I can rewrite the top as 1:

\frac{1}{\cos(x)}

By the quotient identity \sec(x)=\frac{1}{\cos(x)}, I can rewrite this as:

\sec(x)

By the cofunction identity \sec(x)=\csc(x)=(\frac{\pi}{2}-x), we have the second factor of the right hand side:

\csc(\frac{\pi}{2}-x)

Let's just do it all together without all the words now:

\cot(x)+\cot(\frac{\pi}{2}-x)

\cot(x)+\tan(x)

\frac{\cos(x)}{\sin(x)}+\frac{\sin(x)}{\cos(x)}

\frac{1}{\sin(x)}(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)(\cos(x)+\sin(x)\frac{\sin(x)}{\cos(x)})

\csc(x)[\frac{\cos(x)\cos(x)}{\cos(x)}+\sin(x)\frac{sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos(x)\cos(x)+\sin(x)\sin(x)}{\cos(x)}]

\csc(x)[\frac{\cos^2(x)+\sin^2(x)}{\cos(x)}]

\csc(x)[\frac{1}{\cos(x)}]

\csc(x)[\sec(x)]

\csc(x)[\csc(\frac{\pi}{2}-x)]

\csc(x)\csc(\frac{\pi}{2}-x)

7 0
3 years ago
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