Answer:
2 square cm
Step-by-step explanation:
Given :
A square is inscribed in a circle whose radius is r = 1 cm
Therefore, the diameter of the circle is 2 r = 2 x 1
= 2 cm.
So the diagonal of the square is 2r.
Using the Pythagoras theorem, we find each of the side of the triangle is 
.
Therefore, the area of the square is given by 
= 
Hence the area of the largest square that is contained by a circle of radius 1 cm is 2 cm square.
Answer:
$70
Step-by-step explanation:
Middletown Equation:
1c +50 = T
Sweet Shoppe Equation:
2c + 30 = T
Set equations equal to one another and solve for first for cupcakes then for cost.
1c + 50 = 2c + 30
c+ 20 = 2c
20 = c (twenty cupcakes would be ordered)
Now insert into either equation ( or both to check)
1(20) + 50 = 70
2(20)+ 30 = 70
Total cost would be $70
HOPE THIS HELPS!!!
Start from the right-most 7, and go one digit to the left each time:
7 - units place
7 - tens place
7 - hundreds place
3 - thousands place
2 - ten-thousands place
6 - hundred-thousands place
5 - millions place
4 - ten-millions place <----- answer to this question
1 - hundred-millions place
8 - billions place
0 - ten-billions place
9 - hundred billions place
The ten-millions place is the 4.
The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.
Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.
By the fundamental theorem of calculus,
The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so
