Answer: 16 square units
Step-by-step explanation:
let the vertex in quadrant I be (x,y)
then the vertex in quadratnt II is (-x,y)
base of the rectangle = 2x
height of the rectangle = y
Area = xy
= x(12 - x²)
= -x³ + 12x
d(area)/dx = 3x² - 12 = 0 for a maximum of area
3x² = 12
x² = 4
x = ±2
y = 12-4 = 8
So, the largest area = 2 x 8 = 16 square units
The smallest region that can be photographed is
5*7 = 35 square km
so we want to know how long it takes to zoom out to an area of
35 * 5 = 175 square km.
Notice that the length and width increase at the same rate, 2 km/sec, and they start out with a difference of two, so when the width has increased from 5 to x, the length will have increased from 7 to x+2. At the desired coverage area, then,
x(x+2) = 175
x^2 + 2x = 175
x^2 + 2x + 1 = 176 [completing the square]
(x+1)^2 = 176
x+1 = ±4√11
x = -1 ±4√11
Since a negative value for x is meaningless here, x must be
-1 + 4√11 = about 12.27 km
and the time it took to increast to that value was
(12.27 - 5) km / (2 km/sec) = about 3.63 seconds
Answer:
x + y = 50 (1) and 2x + 3y = 115 (2); x = 35, y = 15
Step-by-step explanation:
Since x represents the number of 2 point shots and y represents the number of 3 point shots. If the total number of shots is 50, then x + y = 50 (1)
Also, the total 2 point shots is number of 2 point shots × points per shot = 2x and the total 3 point shots is number of 3 point shots × points per shot = 3y. Since the total number of points is 115, then
2x + 3y = 115 (2)
So, the system of equations are
x + y = 50 (1) and 2x + 3y = 115 (2)
We can solve for x and y by substituting x = 50 - y into (2)
So, 2(50 - y) + 3y = 115
100 - 2y + 3y = 115
100 + y = 115
y = 115 - 100
y = 15
So, x = 50 - y
= 50 - 15
= 35
Answer:
a²+b²=c²
Step-by-step explanation:
a= side of right triangle
b= side of right triangle
c= hypotenuse
Answer:
Step-by-step explanation:
As per midsegment theorem of a trapezoid,
Segment joining the midpoints of the legs of the of the trapezoid is parallel to the bases and measure half of their sum.
Length of midsegment = 
3). MN = 
= 14
4). MN = 
= 66.5
5). MN = 
7 = 
14 = AB + 10
AB = 14 - 10
AB = 4
6). 15 = ![\frac{1}{2}[(3x+2)+(2x-2)]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%283x%2B2%29%2B%282x-2%29%5D)
30 = 5x
x = 6