Take x-2 and insert it into 2x^2 + 3x-2 where the x is located
2x^2 + 3x-2
2(x-2)^2 + 3(x-2)-2
Now work out 2(x-2)^2 + 3(x-2)-2 also follow PEMDAS
2(x-2)^2 + 3(x-2)-2
Since (x-2)^2 is an Exponent, lets work with that first and expand (x-2)^2.
(x-2)^2
(x -2)(x-2)
x^2 -4x + 4
Now Multiply that by 2 because we have that in 2(x-2)^2
(x-2)^2 = x^2 -4x + 4
2(x-2)^2 = 2(x^2 -4x + 4)
2(x^2 -4x + 4) = 2x^2 - 8x + 8
2x^2 - 8x + 8
Now that 2(x-2)^2 is done lets move on to 3(x-2).
Use the distributive property and distribute the 3
3(x-2) = 3x - 6
All that is left is the -2
Now lets put it all together
2(x-2)^2 + 3(x-2)-2
2x^2 - 8x + 8 + 3x - 6 - 2
Now combine all our like terms
2x^2 - 8x + 8 + 3x - 6 - 2
Combine: 2x^2 = 2x^2
Combine: -8x + 3x = -5x
Combine: 8 - 6 - 2 = 0
So all we have left is
2x^2 - 5x
Answer:
The statement is false.
Step-by-step explanation:
If the modeling is with multiple variables it is necessary for all the variables to be modeled well such that the criticality of the model is dependent on all the variograms. It is not dependent on the cross variograms only.
Hello,
7l=7×1000=7000 ml
Bye :-)
This should be easy because we just have to use substitution method. Substitute the value of x from the second equation into the x of the first equation.
-4(2y) + 11y = 15
-8y + 11y = 15
3y = 15 ; y = 5
Substitute this value of y to either the first or second equation.
x = 2(5) = 10
The ordered pair is therefore (10,5).