2m + 11 = 4m + 1
2m (-2m) + 11 (+1) = 4m (+2m) + 1 (-1)
12 = 6m
12/6 = 6m/6
m = 2
hope this helps
Answer:
addition?
Step-by-step explanation:
Let the number of red, black, green and blue balls be R, B, G, U respectively.
B=R (There are as many black balls as red balls)
G+R=10 so G=10-R (Green balls and red balls should add up to 10)
R+10=U (There should be 10 more blue balls than red balls)
The minimal number of balls is 2, 304 so we have the following inequality:
R+B+G+U≥ 2,304
R+(R)+(10-R)+ (R+10)≥2,304
4R≥2,304
R≥2,304/4=576 so R=576
Answer: 576
Answer:
<h3>25m</h3>
Step-by-step explanation:
Perimeter of the rectangular pasture P = 2(L+W)
Area A = LW
L is the length
W is the width
Given
Perimeter = 110m
Area = 750m
If the length of the pasture is 40m longer than the width, then L = W+40
110 = 2L+2w
55 = L+W .....1
750 = LW.....2
Solving simultaneously
from 1; L = 55-W
substitute into 2;
750 = (55-W)W
750 = 55W-W²
-W²+55W -750 = 0
W²-55W+750 = 0
(W²-25W)-(30W+750) = 0
W(W-25)-30(W-25) = 0
(W-25)(W-30) = 0
W-25 = 0 and W-30 = 0
w = 25m and 30m
Since L = 55-W
L = 55-25 = 30m and;
L = 55-30 = 25m
Since we are told that length id longer than the width then, the width we are going to use is 25m
The area of the shaded region in the task content is; 251.2.
<h3>What is the area of the shaded region in the task content?</h3>
It follows from the task content that the area of the shaded region is;
= Area of semicircle IL - Area of semicircle JK + Area of semicircle IJ + Area of semicircle KL.
Hence, since the radius of each semicircle are as follows;
R(IL) = 12
R(IJ) = R(JK) = R(KL) = 4.
Hence, the Area of the shaded region can be evaluated as;
= (π/2)(12² - 4² + 4² +4²).
= π (144 +16)
= (3.14/2) × 160 = 251.2
Read more on Area of semicircle;
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