Basically, we need to find the diagonal of the square. Once we have the diagonal, we need to find if it's more or less than 11 cm, which is diameter of the circle.
This is the formula we'll use:
a²+b²=c²
Since it's a square, a and b are the same.
7²+7²=c²
49+49=c²
98=c²
√98=c
c=9.8995
So, the diagonal is 9.8995
9.8995<11
So, since the diagonal of the square is 9.8995 cm, it is less than the circumference of the circle meaning that it will fit in to the circle without touch the circle's circumference.
Answer:
paper left = 168.56 units²
Step-by-step explanation:
first, figure the area of the square:
area of square = 28x28 = 784 units²
second, figure the area of the circle:
radius = 1/2of 28 = 14
area of circle = Pi x Radius² = (3.14)(14)² = 615.44 units²
subtract area of circle from area of square:
784 - 615.44 = 168.56 units²
The formula for the circumference of a circle is C = 2pi*r, where r is the radius. Here, 7850 units = C = 2pi*r.
Dividing 7850 units by 2pi, we get
7850 units
r = ----------------- = 1249 units (to the nearest unit)
2pi
Answer:
(-1,4)
Step-by-step explanation:
Reflection Rule for over the y axis: A(x,y) A'(-x,y)
Hope this helps!
Answer:
At (-2,0) gradient is -4 ; At (2,0) gradient is 4
Step-by-step explanation:
For this problem, we simply need to take the derivative of the function and evaluate when y = 0 (when crossing the x-axis).
y = x^2 - 4
y' = 2x
The function y = x^2 - 4 cross the x-axis when:
y = x^2 - 4
0 = x^2 - 4
4 = x^2
2 +/- = x
Hence, this curve crosses the x-axis twice, once at (-2,0) and again at (2,0).
The gradient at these points are as follows:
y' = 2(-2) = -4
y' = 2(2) = 4
Cheers.
