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Salsk061 [2.6K]
2 years ago
10

The diameter of a dinner plate is 13 inches. What is the approximate area of the plate to

Mathematics
1 answer:
allsm [11]2 years ago
7 0

Step-by-step explanation:

The formula for the area of a circle is pi x radius squared. So since we know the diameter we can take that and half it giving us radius. Than we plug radius in the formula, so 6.5 squared is 42.25. You than multiply it by pi, and not 3.14 because theirs a possibility of being a couple units off when rounding, and after multiplying it you'll be left with the answer of 132.73

Answer: 132.73

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Find the 114th term of the following sequence
bulgar [2K]

Answer:

456

Step-by-step explanation:

1. Based on the parts of the pattern show you can tell that the term is the step number multiplied by four. 2. Based on that you can multiply step 114 by 4 which gives you 456.

6 0
3 years ago
Help! Click to see question!
EleoNora [17]

Answer:

B.

Step-by-step explanation:

Both equations use x squared

7 0
3 years ago
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Clarence saved 19% of the money he earned. If he earned $90, how much did Clarence save?
Tcecarenko [31]

Answer:

probably like 10

Step-by-step explanation:

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
You start a savings account and on the first week you deposit $2. Every
vlada-n [284]

Answer:

558 weeks

Step-by-step explanation:

let x = number of weeks

this equation can be derived from the question

2 + 1x = 560

collect like terms

x = 560 - 2

x = 558

5 0
3 years ago
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