Answer:
80 inches
Step-by-step explanation:
6 x 12 = 72 + 8= 80
Give brainliest please
The answer is fifteen yards
It is the third equation: −512⋅(65⋅13)⋅92=(−512⋅65)⋅(13⋅92)
Associative property means one can group the individual operations arbitrarily without changing the result.
Answer:
- 2(L +W) ≤ 600
- W ≤ 200
- L ≥ 2W
Step-by-step explanation:
We assume the problem wording means the length is to be at least 2 times <em>as long as</em> the width. (<em>Longer than</em> usually refers to a difference, not a scale factor.)
If we let "W" and "L" represent the width and length, respectively, then we can translate the problem statement to ...
2(L + W) ≤ 600 . . . . . . the perimeter is twice the sum of length and width
W ≤ 200 . . . . . . . . . . . . the width is at most 200 inches
L ≥ 2W . . . . . . . . . . . . . the length is at least twice the width
Answer:
1. 200 in.^2
2.
Step-by-step explanation:
1. Area of a trapezium=1/2(a+b)height
a=14 in.
b=20 in.+6 in. = 26 in.
height= ?
height can be gotten by using
formula : Soh Cah Toa
Toa is tangent∆= opposite/ adjacent
∆ = 60°
opp=? adj=6 in.
Tan 60° =opp/6 in.
Tan 60° = 1.732
1.732= opp/ 6 in.
opp=1.732*6 in.
opp = 10.392 approximately= 10 in.
Area of trapezium= 1/2*(14+26)*10(in.^2)
Area of trapezium= 1/2*40*10
= 1/2*400
= 200 in.^2 .
2. Area of triangle= 1/2* base* height
for first triangle =
base=4m
height =15m
Area: 1/2*4*15 ,1/2*60= 30m^2
second triangle =
base:?
height: 15m
using Pythagorean theorem;
hyp^2 = opposite ^2 + adjacent ^2
17^2 = 15^2 + adj^2
289= 225 + adj^2
adj^2 = 289-225= 64
square root
adj = 8m
area of second triangle : 1/2*8*15
: 1/2*120 :60m^2
total area: 30 m^2 +60m^2
==90 m^2
