When x = 3 and y = 5, by how much does the value of 3x^2 – 2y exceed the value of 2x^2– 3y ?
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calendarEducator since 2010
write12,551 answers
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When the value of x is equal to 3 and the y is equal to 5,
3x^2 – 2y
=> 3*3^2 - 2*5
=> 3*9 - 10 = 17
And 2x^2– 3y
=> 2*3^2 - 3*5
=> 2* 9 - 15 = 3
This gives 3x^2 – 2y exceeding 2x^2– 3y by 17 - 3 = 14
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
Given the equations:
3x^2 - 2y
2x^2 - 3y
Also we know that x= 3 and y= 5
We need to know the difference in values between both equations.
First we will substitute x and y values in each equation.
==> 3x^2 -2y = 3*(3^2) - 2*5 = 27 -10= 17
==> 2x^2 -3y = 2*(3^2) - 3*5 = 18-15 = 3
Now we will calculate the difference between both values.
==> 17 -3 = 14.
Then the values of 3x^2 -2y exceeds the values of 2x^2 -3y by 14.