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Brut [27]
3 years ago
15

Is square root of 12 times the square root of 3 rational or irrational

Mathematics
1 answer:
TEA [102]3 years ago
7 0

Answer:

Irrational

Step-by-step explanation:

Square root if 12 is 3.464 and the square root of 3 is 1.732. You multiply 3.464 and 1.732 to get 5.99 which would be 6 if you round it up. The reason to why it's irrational is because 6 can not be written as a ratio of two integers, which means it cannot be expressed as a fraction.

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suter [353]

Answer:

1) Since AP/AB=PQ/BC, 6x/(3x+5)=x/1. Cross multiply the denominator and you will get 6x=(3x+5)*x. Simplifying it more and you are left with 3x^2-x=0

2) 3x^2-x=0. x(3x-1)=0. x=0 or x=1/3 but x=0 can't be a solution since it would make the figure absurd. x=1/3 cm

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3 years ago
2y = 2x-4<br> y + 3x = 10<br><br> (substitution)
devlian [24]

Answer:

x = 3, y = 1

Step-by-step explanation:

2y = 2x - 4, so y = x - 2. Substituting this into the bottom equation and solving for x:

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4 0
1 year ago
Pre-Calculus - Systems of Equations with 3 Variables please show work/steps
inessss [21]

Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

x + 0 y - 3 z = 7 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y/2 - 2 z = 3/2 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10 , y = -7 , z = 1

3 0
3 years ago
2. Which of the following (-6, 20), (-9, 1) or (14, 2) ) are a solution to the system of equations?: 2x + y = 8 5x + 2y=10
Leya [2.2K]
(-6, 20)

You can get this through elimination by multiplying the first equation by -2 to get the y values to cancel out. 
8 0
3 years ago
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