Plug in -2 for a, 3 for b, and -5 for c. Then it'd be | (-3)^2 - 2(-2)(-5) +5 (3) |. Then, multiply everything to get | 9 -20 +15 |. Then, add/subtract to get | 4 |. Finally, take the absolute value and get 4.
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
<h3>
Answer: 29.5 degrees</h3>
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Explanation:
Add up the arc measures and divide by 2 to get the angle formed by the two chords.
angle AOB = (arc AB + arc CD)/2
angle AOB = (34 + 25)/2
angle AOB = 59/2
angle AOB = 29.5 degrees
Answer:
The answer to your question is 113.04 in²
Step-by-step explanation:
To solve this problem substitute the values in the formula and simplify.
Data
Area = ?
radius = 6 in
π = 3.14
Formula
Area = π x r²
Substitution
Area = (3.14) x (6)²
Simplification
Area = 3.14 x 36
Result
Area = 113.04 in²
The interval that f(x) is increasing is the distance from 200 to 300.
The minimum value of f(x) in the interval 0<x<300 is 200.
At a value of 500, the value of f(x) is 0.
The function can't be a quadratic function since there are two points in the graph where f(x) changes its rate from increasing to decreasing or the opposite. A quadratic function has only one of that point.
