Answer:
Step-by-step explanation:
P+29=³√x
Find the cube of both side
(P+29)³=(³√x)³
(P+29)³=x
Answer:
Step-by-step explanation:
What this question is asking of you is what is the greatest common divisor of 12 and 15. Or, what is the biggest number that divides both 12 and 15.
in order to find this we have to split each number into it's prime components.
for 12 they are 2,2 and 3 (
2
⋅
2
⋅
3
=
12
)
and for 15 they are 3 and 5 (
3
⋅
5
=
15
)
Out of those two groups (2,2,3) and (3,5) the only thing in common is 3, so 3 is the greatest common divisor. That tells us that the greatest number of groups that can exist and have the same number of girls and the same number of boys for each group is 3.
Now to find out how many girls and boys there are going to be in each group we divide the totals by 3, so:
12
3
=
4
girls per group, and
15
3
=
5
boys per group.
(just as a thought exercise, if there were 16 boys, the divisors would have been (2,2,3) and (2,2,2,2), leaving us with 4 groups [
2
⋅
2
] of 3 girls [12/4] and 4 boys [16/4] )
By inspecting the integrand, the "obvious" choice for substitution would be
<em>u</em> = <em>y</em> + <em>x</em>
<em>v</em> = <em>y</em> - <em>x</em>
<em />
Solving for <em>x</em> and <em>y</em>, we would have
<em>x</em> = (<em>u</em> - <em>v</em>)/2
<em>y</em> = (<em>u</em> + <em>v</em>)/2
in which case the Jacobian and its determinant are

The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.
Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So
-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>
-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2
-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>
-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>
<em />
So, the integral becomes



