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Lapatulllka [165]
3 years ago
10

Which explanation justifies how the area of a sector of a circle is derived?

Mathematics
1 answer:
givi [52]3 years ago
4 0
The answer would be D
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Why i the product of two rational number alway rational?
ella [17]

The product of two rational numbers is always rational because (ac/bd​) is the ratio of two integers, making it a rational number.

We need to prove that the product of two rational numbers is always rational. A rational number is a number that can be stated as the quotient or fraction of two integers : a numerator and a non-zero denominator.

Let us consider two rational numbers, a/b and c/d. The variables "a", "b", "c", and "d" all represent integers. The denominators "b" and "d" are non-zero. Let the product of these two rational numbers be represented by "P".

P = (a/b)×(c/d)

P = (a×c)/(b×d)

The numerator is again an integer. The denominator is also a non-zero integer. Hence, the product is a rational number.

learn more about of rational numbers here

brainly.com/question/29407966

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8 0
1 year ago
Please, help,,,,, thanks
Veseljchak [2.6K]
Is this correct ? Hope this helped!

7 0
3 years ago
[2 x 2(2 + 1)^2= please help
aleksandrvk [35]

Answer:

36x

Step-by-step explanation:

PEMDAS RULE

3 0
3 years ago
HELP ASAP 30 POINTS WORTH !!!!! OωO
Len [333]

Answer:

A because (0,-6) is the y-intercept so you start at the point you know. Then because slope is a rise over run fraction the slope can also be written as 2/1 which is rise 2 over 1.

Step-by-step explanation:

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6 0
2 years ago
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Please help: linear algebra problem. (Linear combinations)
DochEvi [55]

Answer:

\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

This tells us that:

A=\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right]

b=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

Step-by-step explanation:

So we are saying we have scalars, c and d, such that:

c\left[\begin{array}{ccc}5\\5\\ 3\end{array}\right]+d\left[\begin{array}{ccc}7\\-8\\-9\end{array}\right]=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right].

So we want to find a way to express this as:

Ax=b where x is the scalar vector, \left[\begin{array}{ccc}c\\d\end{array}\right].

So we can write this as:

\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

3 0
3 years ago
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