Answer:
Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle.
Step-by-step explanation:
Ok, so first we distribute, you multiply the seven into everything in the parentheses next to it. So far we have, 14a+21+3(4a-2). You distribute the three into the parentheses to get, 14a+21+12a-6. You combine the like terms to get, 26a-15. You cannot simplify it any further so the answer is 26a-15.
Answer:
9
Step-by-step explanation:
forget all this grammar,
let the number be x,
x = {18(15-12)} / 6
x = {18(3)} / 6
x= 54/6
x= 9
value is therefore 9
Answer: 5.6
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Work Shown:
cos(angle) = adjacent/hypotenuse
cos(B) = AB/BC
cos(62) = AB/12
12*cos(62) = AB
AB = 12*cos(62)
AB = 5.63365875
AB = 5.6
Simplify the integrands by polynomial division.
Now computing the integrals is trivial.
5.
where we use the power rule,
and a substitution to integrate the last term,
8.
using the same approach as above.
