The students in math class use square tiles to make arrays. If they
- use 8 (8=1·2·2·2) tiles, then they can form arrays of the length in 1 tile, 2 tiles, 4 tiles and 8 tiles;
- use 9 (9=1·3·3) tiles, they can form arrays of the length in 1 tile, 3 tiles and 9 tiles (one array less).
So, you can conclude that Celia is correct.
Answer:
V=33.51in^3
Step-by-step explanation:
5inch-1 in
=4inch
divide by 2 because the problem gives diameter, not radius. (radius is half of diameter)
The space inside is a sphere so it will be
V=4/3πr^3
V=4/3*pi*(2)^3
V=33.51 inch cubed
Answer:
P = 57°
Step-by-step explanation:
Given the following :
PQ = 17
QR = 15
PR = 14
Using the cosine formula since the length of the three sides are given:
a2 = b2 + c2 – 2bccos(A)
To find P:
QR^2 = PQ^2 + PR^2 – 2(PQ)(PR)cos(P)
15^2 = 17^2 + 14^2 – 2(17)(14)cos(P)
225 = 289 + 196 - 476 cosP
476*CosP = 485 - 225
476*CosP = 260
CosP = 260/476
CosP = 0.5462184
P = Cos^-1(0.5462184)
P = 56.892029
P = 57°
Answer: x = 70, y = 110, z = 85
Step-by-step explanation:
If BE = CD, then using corresponding angles (are equal), x = 70
If x = 70, then 70+y = 180 (angles on a straight line)
y = 110
And z = 85 (corresponding angles again): angle ABE = angle ACD