3.875 Should be adding to 6.125 to get 10, however if you want an easier way of checking this. Do 3.875 + 6.125 and you'll get exactly 10. Click thanks <3

4 0

2 years ago

Solve the following question

nydimaria [60]

9514 1404 393

Answer:

π/3

Step-by-step explanation:

The given integral does not exist. We assume there is a typo in the upper limit, and that you want the integral whose upper limit is (√3)/2.

It is convenient to make the substitution ...

x = sin(y) . . . . so, y = arcsin(x)

dx = cos(y)·dy

Then the integral is ...

$\displaystyle\int_0^{\sqrt{3}/2}{\dfrac{\cos{y}}{\sqrt{1-\sin^2{y}}}}\,dy=\int_0^{\sqrt{3}/2}{dy}=\left.\arcsin{x}\right|\limits_0^{\sqrt{3}/2}\\\\=\arcsin{(\sqrt{3}/2)}=\boxed{\dfrac{\pi}{3}}$

5 0

2 years ago

Find the area. The figure is not drawn to scale. 1. 36 in. 40 in. 33 in. -

mote1985 [20]

Answer: 47,520

Step-by-step explanation: 36 times 40 times 33

5 0

2 years ago