The picture in the attached figure
we know that
Area of circle=pi*r²
area of the shaded region=364*pi cm²
area of the shaded region=area of the outer circle -area of the<span> inside circle
</span>area of the outer circle=pi*[2x]²----> 4*x²*pi cm²
area of the inside circle=pi*6²------> 36*pi cm²
364*pi=4*x²*pi-36*pi------> 364=4*x²-36
4*x²=364+36-----> 4*x²=400
x²=400/4----> x²=100
x=10 cm
the answer isx=10 cm
We will use the law of cosines
<span>side a² = b² + c² -2bc • cos(A)
</span><span>side a² = 729 + 196 -2*27*14 * cos (46)
</span><span>side a² = 925 -(756 * 0.69466)
</span>side a² = <span><span>399.83704
</span>
side a = </span><span><span><span>19.995925585
</span>
</span>
</span>
We could round that to 20
a = 20 b = 27 c =14
We can calculate a triangle's area when we know all 3 sides by using Heron's Formula
<span>area = square root (s • (s - a) • (s - b) • (s - c))
where s is the semi-perimeter </span>
semi-perimeter<span> = (side a + side b + side c) ÷ 2</span>
s = (20 + 27 + 14) / 2
s = 30.5
Now we use Heron's Formula
area = square root (s • (s - a) • (s - b) • (s - c))
area = square root (30.5 • (<span>30.5 - 20) • (</span><span>30.5 - 27) • (</span><span>30.5 - 14))</span>
area = square root (30.5 • (10.5) • (3.5) • (<span>16.5))</span>
<span>area = square root (18494.4375)
</span>
<span><span><span>area = 135.9942553934
</span>
</span>
</span>which rounds to
136 square feet
Source:
http://www.1728.org/triang.htm
Answer:
The correct answer is Option A.) The standard deviation of the distribution of sample means is less than the population standard deviation of the number of cancer spots.
Step-by-step explanation:
The standard deviation of the distribution of sample means is less than the population standard deviation of the number of cancer spots.
Answer:
No
Step-by-step explanation:
The way to find the line of best fit by estimate is to have about half the points be above and below the line of best fit. In this case Tariq followed the first few points of the data but his estimate would be very off after 10 on the x axis. This would not accurately predict what the next data point could be.
First of all, a bit of theory: since the area of a square is given by
where s is the length of the square. So, if we invert this function we have
.
Moreover, the diagonal of a square cuts the square in two isosceles right triangles, whose legs are the sides, so the diagonal is the hypothenuse and it can be found by
So, the diagonal is the side length, multiplied by the square root of 2.
With that being said, your function could be something like this:
double diagonalFromArea(double area) {
double side = Math.sqrt(area);
double diagonal = side * Math.sqrt(2);
return diagonal;
}
