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Andreyy89
4 years ago
9

What is the following product? (√12+√6)(√6-√10)

Mathematics
2 answers:
olasank [31]4 years ago
6 0
<span>(√12+√6)(√6-√10)
= </span>√12 √6 - √12 √10 +√6 √6  - √6√10
= √72 - √120 + 6 - √60
= 6√2 - 2√30 + 6 - 2√15

Answer is A
6√2 - 2√30 + 6 - 2√15
koban [17]4 years ago
4 0
ANSWER

6 \sqrt{2} -2 \sqrt{30}  +6 -  2\sqrt{15}

EXPLANATION

We want to find the product

( \sqrt{12}  +  \sqrt{6} )( \sqrt{6} -  \sqrt{10}  )

We apply the distributive property to get,


\sqrt{12} ( \sqrt{6} -  \sqrt{10} ) +\sqrt{6} ( \sqrt{6} -  \sqrt{10} )



We expand to obtain,

\sqrt{12}  \times \sqrt{6} - \sqrt{12}   \times \sqrt{10}  + \sqrt{6}  \times  \sqrt{6}  -   \sqrt{6} \times  \sqrt{10}


We now simplify to get,


6 \sqrt{2} -2 \sqrt{30}  +6 -  2\sqrt{15}

The correct answer is option A.
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A factory makes 1200 shirts every 6 hours the factory makes shirts for 9 hours each workday enter the fewest number of workdays
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Hey make 200 per hour, and would need 63 hours to make 12600 shirts. 63/9=7
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4 years ago
Read 2 more answers
Tell whether the lines through the given points are parallel, perpendicular, or neither. Explain.
mina [271]

Answer:

parallel

Step-by-step explanation:

find the slope of each line

Line 1   (5-2)/(-4-2) = 3/-6 =-1/2

Line 2  (-4--9)/(-6-4)= 5/-10 = -1/2

5 0
3 years ago
a square poster of length 3x is going to have a square painting centered on it. The length of the painting is 2x. the area of th
eduard
One way to solve this is to look at the size of the poster, and then subtract the size of the picture so you have the area that needs to be painted left.

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3 0
4 years ago
In a right triangle ΔABC, the length of leg AC = 5 ft and the hypotenuse AB = 13 ft. Find: Chapter Reference b The length of the
Butoxors [25]

Answer:

The length of the angle bisector of angle ∠A is 6.01.

Step-by-step explanation:

It is given that length of leg AC = 5 ft and the hypotenuse AB = 13 ft.

Using pythagoras theorem

(AB)^2=(BC)^2+(AC)^2

(13)^2=(BC)^2+(5)^2

169=(BC)^2+25

BC=12

\sin A=\frac{\text{perpendicular}}{\text{hypotenuse}}

\sin A=\frac{BC}{AB}

A=\sin ^{-1}\frac{12}{13}

A=67.38

Bisector divides the angle in two equal parts, therefore,

A'=\frac{67.38}{2} =33.69

In triangle ACD.

\cos A'=\frac{\text{Base}}{\text{Hypotenuse}}

\cos A'=\frac{AC}{AD}

\cos (33.69^{\circ})=\frac{5}{AD}

0.832=\frac{5}{AD}

AD=\frac{5}{0.832} =6.009\approx 6.01

Therefore the length of the angle bisector of angle ∠A is 6.01.

4 0
3 years ago
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