Answer:
a. 6Pack of paint brushes for $12.99
b. 5 classes for $64.99
c. 12 pack of paper plates
Step-by-step explanation:
a.
I could buy a 6 pack of paint brushes for $ 12.99
or
buy 6 individual brushes for the price of $ 2.19 each.
6 x $2.19 = $13.14
first option is cheaper
b.
10 classes for s total of $150
$150 ÷ 10 = $15 (price per class)
5 classes for $64.99
$64.99 ÷ 5 = $12.998 (price per class)
c.
12 pack of paper plates for $2.29
25 pack for $4.99
$2.29 ÷ 12 = $0.1908 (price of one plate)
$4.99 ÷ 25 = $0.1996 (price for one plate)
The most probable number, in this case, would be by ratio,
32:48 = 360:x
cross multiply and solve for x
x=360*48/32=540
Ans. Most probably there are 540 students enrolled.
Answer:
m ∠ W is 58.2°
Step-by-step explanation:
Given:
XY = 15
WZ = 22
∠ XZY =25°
To Find:
m ∠ W = ?
Solution:
In right angle triangle Δ XZY, by Sine identity we have
![\sin Z= \frac{\textrm{side opposite to angle Z}}{Hypotenuse} \\\sin 25= \frac{XY}{XZ}\\0.4226=\frac{15}{XZ}\\\therefore XZ=\frac{15}{0.4226} \\XZ=35.49](https://tex.z-dn.net/?f=%5Csin%20Z%3D%20%5Cfrac%7B%5Ctextrm%7Bside%20opposite%20to%20angle%20Z%7D%7D%7BHypotenuse%7D%20%5C%5C%5Csin%2025%3D%20%5Cfrac%7BXY%7D%7BXZ%7D%5C%5C0.4226%3D%5Cfrac%7B15%7D%7BXZ%7D%5C%5C%5Ctherefore%20XZ%3D%5Cfrac%7B15%7D%7B0.4226%7D%20%5C%5CXZ%3D35.49)
∴ XZ = 35.49
Now in right angle triangle Δ WZX, by tangent identity we have
![\tan W = \frac{\textrm{side opposite to angle Z}}{\textrm{side adjacent to angle Z}}\\\tan W = \frac{XZ}{WZ}\\\tan W = \frac{35.49}{22}\\\tan W = 1.6131\\\therefore W =\tan^{-1}(1.6131) \\W= 58.2\°](https://tex.z-dn.net/?f=%5Ctan%20W%20%3D%20%5Cfrac%7B%5Ctextrm%7Bside%20opposite%20to%20angle%20Z%7D%7D%7B%5Ctextrm%7Bside%20adjacent%20to%20angle%20Z%7D%7D%5C%5C%5Ctan%20W%20%3D%20%5Cfrac%7BXZ%7D%7BWZ%7D%5C%5C%5Ctan%20W%20%3D%20%5Cfrac%7B35.49%7D%7B22%7D%5C%5C%5Ctan%20W%20%3D%201.6131%5C%5C%5Ctherefore%20W%20%3D%5Ctan%5E%7B-1%7D%281.6131%29%20%5C%5CW%3D%2058.2%5C%C2%B0)
∴m ∠W =58.2°
The sides are 2, 1, 1.5, and 2.5 clockwise
Cavalieri's Principle states that for two different solid shapes, if the altitude of the shapes is equal and the cross sections these shapes yield are the same from equal distances from their bases, then the shapes have equal volume.
Simply stated, if you cut two shapes of equal height from the same spot and they repeatedly yield the same cross section, then the shapes have equal volume. This is true in the case of a regular cylinder and an oblique cylinder. So the volume of the oblique cylinder may be calculated using πr²h, since its volume is equivalent to a normal cylinder of the same dimensions.
Volume = π x 10² x 20
Volume = 2000π cm³