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ch4aika [34]
3 years ago
7

Point R(3,5) is rotated 90 degrees clockwise about the origin. What are the coordinates of R'?

Mathematics
1 answer:
maria [59]3 years ago
3 0

Answer:

(5,-3)

Step-by-step explanation:

if this is the rotation that I remember learning, it should be correct. I'm very sorry if it isn't right

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What is the answer to this ? <br><br> 1.39×2
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The answer is 2.78

Step-by-step explanation:

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If somebody is writing a book and the number of hours they are writing is 5:8 how many pages each hour did they write
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They write 5 pages per 8 hours
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Henry is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices. Comp
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Step-by-step explanation:

step by step explanation

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3 years ago
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1. What types of numbers make up the set of rational numbers?
Ad libitum [116K]

Answer:

Natural Numbers, Whole Numbers, and Integers are rational numbers.

Example: {1, 2, 3, -1, -2, -3}

Numbers with repeating decimals and terminating decimals are also rational.

Example: {2.34343434..., 3.45}

If a square root makes a perfect square, then it is rational.

Example: {√16, √100}

Hope this helps.

8 0
3 years ago
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
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