Answer:
The percent error of the measurement, to the nearest tenth of a percent is 
Step-by-step explanation:
Estimated length of a line = 12.4 inches
Actual length of the line = 15.7 inches
Error = 15.7 - 12.4 = 3.3 inches
Percent error of the measurement 
≈ 21.0
So, the percent error of the measurement, to the nearest tenth of a percent is
That figure is made up of 3 shapes.
A square, a rectangle, a semicircle.
The square has a side of 6 ft.
The semicircle has a radius of 4 ft.
The rectangle's width is the diameter of the semicircle, so it is 8 ft. The length of the rectangle is 18 ft.
Square:
A = s^2 = (6 ft)^2 = 36 ft^2
Rectangle:
A = LW = 18 ft * 8 ft = 144 ft^2
Semicircle:
A = (1/2) * pi * r^2 = (1/2) * pi * (4 ft)^2 = (1/2) * pi * 16 ft^2 = 8pi ft^2
Total area:
A = 36 ft^2 + 144 ft^2 + 8pi ft^2
A = (180 + 8pi) ft^2
Answer:
40 feet
Step-by-step explanation:
We know that the area of a rectangle is represented as 
, where l is the length and w is the width.
We already know the width, and we know the area, so we can plug these values into the equation.
Our goal is to now isolate the variable l, and to do this we can divide both sides by 6.
Hope this helped!
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These are similarity criterion
<span>X^2+x-2/1-x^2 ÷ 2x^2-6x-20/x^2-4x-5
(x + 2)(x - 1) (x - 5)(x + 1)
= ------------------- * ---------------------
(1 + x)(1- x) 2(x + 2)(x - 5)
</span> (x - 1)
= ----------
2(1- x)
- (1 - x)
= -------------
2(1 - x)
= -1/2
answer
-1/2
