Answer:
A,C,D
...............................
Looking at this problem in the book, I'm guessing that you've been
introduced to a little bit of trigonometry. Or at least you've seen the
definitions of the trig functions of angles.
Do you remember the definition of either the sine or the cosine of an angle ?
In a right triangle, the sine of an acute angle is (opposite side) / (hypotenuse),
and the cosine of an acute angle is (adjacent side) / (hypotenuse).
Maybe you could use one of these to solve this problem, but first you'd need to
make sure that this is a right triangle.
Let's see . . . all three angles in any triangle always add up to 180 degrees.
We know two of the angles in this triangle ... 39 and 51 degrees.
How many degrees are left over for the third angle ?
180 - (39 + 51) = 180 - (90) = 90 degrees for the third angle.
It's a right triangle ! yay ! We can use sine or cosine if we want to.
Let's use the 51° angle.
The cosine of any angle is (adjacent side) / (hypotenuse) .
'BC' is the side adjacent to the 51° angle in the picture,
and the hypotenuse is 27 .
cosine(51°) = (side BC) / 27
Multiply each side of that equation by 27 :
Side-BC = (27) times cosine(51°)
Look up the cosine of 51° in a book or on your calculator.
Cosine(51°) = 0.62932 (rounded)
<u>Side BC</u> = (27) x (0.62932) = <u>16.992</u> (rounded)
============================================
You could just as easily have used the sine of 39° .
That would be (opposite side) / (hypotenuse) ... also (side-BC) / 27 .
The long edge of the paper is 11 inches, so you would take 4 papersx11 inches.
4x11=44
The long edge of the frame is 44 inches.
Answer:
![A_{f}=4\pi (\sqrt[3]{36} r)^{2}\\\\V_{f}=\frac{4}{3} \pi (36r^{3})](https://tex.z-dn.net/?f=A_%7Bf%7D%3D4%5Cpi%20%28%5Csqrt%5B3%5D%7B36%7D%20r%29%5E%7B2%7D%5C%5C%5C%5CV_%7Bf%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%2836r%5E%7B3%7D%29)
Step-by-step explanation:
In order to find the final radii of the sphere, we need to calculate the volume, knowing that volumes are additive:
![V_{1}=\frac{4}{3} \pi (r^{3})\\\\V_{2}=\frac{4}{3} \pi (2r)^{3}\\\\V_{3}=\frac{4}{3} \pi (3r)^{3}\\\\V_{f}=\frac{4}{3} \pi (r^{3}+(2r)^{3}+(3r)^{3})\\\\V_{f}=\frac{4}{3} \pi (r^{3}+8r^{3}+27r^{3})\\\\V_{f}=\frac{4}{3} \pi (36r^{3})\\\\V_{f}=\frac{4}{3} \pi R^{3}\\\\R=\sqrt[3]{36} r](https://tex.z-dn.net/?f=V_%7B1%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%28r%5E%7B3%7D%29%5C%5C%5C%5CV_%7B2%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%282r%29%5E%7B3%7D%5C%5C%5C%5CV_%7B3%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%283r%29%5E%7B3%7D%5C%5C%5C%5CV_%7Bf%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%28r%5E%7B3%7D%2B%282r%29%5E%7B3%7D%2B%283r%29%5E%7B3%7D%29%5C%5C%5C%5CV_%7Bf%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%28r%5E%7B3%7D%2B8r%5E%7B3%7D%2B27r%5E%7B3%7D%29%5C%5C%5C%5CV_%7Bf%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%2836r%5E%7B3%7D%29%5C%5C%5C%5CV_%7Bf%7D%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20R%5E%7B3%7D%5C%5C%5C%5CR%3D%5Csqrt%5B3%5D%7B36%7D%20r)
Now that we know the radii of the new sphere, we can calculate the surface area:
![A_{f}=4\pi R^{2}\\\\A_{f}=4\pi (\sqrt[3]{36} r)^{2}](https://tex.z-dn.net/?f=A_%7Bf%7D%3D4%5Cpi%20R%5E%7B2%7D%5C%5C%5C%5CA_%7Bf%7D%3D4%5Cpi%20%28%5Csqrt%5B3%5D%7B36%7D%20r%29%5E%7B2%7D)
The second one the are both perfect squares
