1. 5 in and 1/3 in: Area = 5/3 in^2
2. 5 in and 4/3 in: Area= 20/3 in^2
3. 5/2 in and 4/3 in: Area=10/3 in^2
4. 7/6 in and 6/7 in: Area = 1 in^2
Step-by-step explanation:
<u>1. 5 in and 1/3 in</u>
Here,
<u>2. 5 in and 4/3 in</u>
Here,
<u>3. 5/2 in and 4/3 in</u>
<u>4. 7/6 in and 6/7 in</u>
<u>Let</u>
<u>Hence,</u>
1. 5 in and 1/3 in: Area = 5/3 in^2
2. 5 in and 4/3 in: Area= 20/3 in^2
3. 5/2 in and 4/3 in: Area=10/3 in^2
4. 7/6 in and 6/7 in: Area = 1 in^2
Keywords: Rectangle, Area
Learn more about rectangles at:
#LearnwithBrainly
Answer:
h=32 cm
Step-by-step explanation:
12*2=24
+8=32
b (?) and c
m and n have the same slope (-2/5) so m//n
Let W = width of package
Let H = height of package
Let L = length of package
The perimeter cab be one of the following:
P = 2(L + W), or
P = 2(L + H)
The perimeter of the cross section cannot exceed 108 in.
When the width is 10 in, then
2(L + 10) <= 108
L + 10 <= 54
L <= 44 in
When the height is 15 in, then
2(L + 15) <= 108
L + 15 <= 54
L <= 39 in
To satisfy both of these conditions requires that L <= 39 in.
Answer: 39 inches
Answer:
3a. 8t + 12c = 72
3b. t + c = 7
Step-by-step explanation:
number of tapes is t
number of Cds is c
so 8t + 12c = 72
and t + c = 7
so t = 7 - c
replace t = 7 - c in to the first equation
8(7 - c) + 12c = 72
56 - 8c + 12c = 72
4c = 72 - 56 = 16
c = 4
if c = 4, then t = 7 - c = 7 - 4 = 3
