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3 years ago

Question #2 How many different groups of 10 acts can be selected to play in the winter show? *

Mathematics

1 answer:

Marizza181 [45]3 years ago

4 0

Answer:

1001

Step-by-step explanation:

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What is cramers rule.

Jobisdone [24]

Cramer's Rule is a method that uses determinants to solve systems of equations that have the same number of equations as variables

6 0

3 years ago

Simplify the expression below:

kifflom [539]

Answer:

C. -7x^2 - 28

Step-by-step explanation:

Firstly simplify the bracket before doing anything;

4x(3x - 7) = 12x^2 - 28x

then go back to the question and substitute the expression(12x^2 - 28x) on the bracket and then work out the question;

12x^2 - 28x -19x^2...then group the values with the same exponent of x,

(12x^2 - 19x^2) - 28x

; -7x^2 - 28x

6 0

3 years ago

In a triangle a=20 and c=29 but how long is side b ?

Afina-wow [57]

A^2+b^2=c^2
20^2+b^2=29^2
400+b^2=841
-400. -400
b^2=√441
b=21

6 0

3 years ago

Match the following rational expressions to their rewritten forms.

Nookie1986 [14]

Answer:

Answer image is attached.

Step-by-step explanation:

Given rational expressions:

$1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}$

And the rewritten forms:

$(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}$

We have to match the rewritten terms with the given expressions.

Let us consider the rewritten terms and let us solve them one by one by taking LCM.

$(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}$

So, correct option is 3.

$(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}$

So, correct option is 1.

$(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}$

So, correct option is 2.

$(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}$

So, correct option is 4.

The answer is also attached in the answer area.

7 0

3 years ago

Solve for the solution: (x-4)^2-4=0

kondor19780726 [428]

Step-by-step Explanation:

x=6,x=2

5 0

3 years ago