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joja [24]
3 years ago
12

When x =6 y=2 What is: 2(3x-6y)^2

Mathematics
2 answers:
sashaice [31]3 years ago
6 0

Answer:

2(3x - 6y) ^{2}  = 72

Step-by-step explanation:

2(3x - 6y) ^{2}

x = 6 \:  \: y = 2

Substitute in the values

2(3(6) - 6(2))^{2}

2(18 - 12) ^{2}

2(6) ^{2}

2(36)

= 72

ikadub [295]3 years ago
3 0

Answer:

72

Step-by-step explanation:

We first substitute x = 6 and y = 2 into the equation

2(3 x (6) - 6 x (2)) ^ 2

First we multiply all the numbers in the bracket

2(18 - 12) ^ 2

Now we minus the numbers in the bracket

2(6)^2

Now we use the indice rule to do 6^2 which is 36

And then we multiply 36 by 2 which is 72

Hope it helps :)

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(3x-70) <br> (3y+40)<br> 120<br> x
Andru [333]
(3y+40)+(3x-70)=180

3y+3x-30=180
3y+3x=210
3(y+x)=210
y+x=70
x=70-y

meanwhile 180-120=x therefore x=60
therefore
60=70-y y=10
so x=60 y=10
8 0
3 years ago
An angle measures 62.4° more than the measure of its supplementary angle. What is the measure of each angle?
Klio2033 [76]

Answer:

Step-by-step explanation:

If an angle measures x degrees, the measure of its supplement is (180-x) degrees.

This means that x-62.4 = 180-x, meaning that x=121.2.

So, the angles are 121.2 and 58.8 degrees.

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2 years ago
The domain of F(x) is the set of all numbers greater than or equal to 0 and less than it equal to 2
antoniya [11.8K]
Its false cause its not the set of all numbers
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3 years ago
What is the square root of 500
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3 0
3 years ago
Boris choos three diffrent numbers.the sum of the three numbers is 36.One of trh numbers is a cube number.the other two number a
inna [77]

Answer:

4,5,27

Problem:

Boris chose three different numbers.

The sum of the three numbers is 36.

One of the numbers is a perfect cube.

The other two numbers are factors of 20.

Step-by-step explanation:

Let's pretend those numbers are:

a,b, \text{ and } c.

We are given the sum is 36: a+b+c=36.

One of our numbers is a perfect cube. a=n^3 where n is an integer.

The other two numbers are factors of 20. bk=20 and ci=20 where a,c,i, \text{ and } k \text{ are integers}.

n^3+\frac{20}{k}+\frac{20}{i}=36

From here I would just try to find numbers that satisfy the conditions using trial and error.

3^3+\frac{20}{2}+\frac{20}{2}

27+10+10

47

3^3+\frac{20}{4}+\frac{20}{5}

27+5+4

36

So I have found a triple that works:

27,5,4

The numbers in ascending order is:

4,5,27

4 0
3 years ago
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