A²+b²=c²
a=b because it has 2 45° angles and one 90 so it is isosceles right triangle
∴ a² + a² = c²
2a²=24²
2a²=576
a²=288
a=√288
a=16.97056275
Therefore. With A=1/2bh
A=(1/2) * 16.97056275 * 16.97056275
= (1/2) * 288
= 144
144 ft²
In place of t, or theta, I'm going to utilize x instead. So the equation is -3*cos(x) = 1. Get everything to one side and we have -3*cos(x)-1 = 0
Let f(x) = -3*cos(x)-1. The goal is to find the root of f(x) in the interval [0, 2pi]
I'm using the program GeoGebra to get the task done of finding the roots. In this case, there are 2 roots and they are marked by the points A and B in the attachment shown
A = (1.91, 0)
B = (4.37, 0)
So the two solutions for theta are
theta = 1.91 radians
theta = 4.37 radians
-2x + 6y = 6
6y - 6 = 2x
6(y-1) = 2x
(1/2) * 6(y-1) = 2x * (1/2)
x = 3(y-1)
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If, x = 3(y-1),
-7 * 3(y-1) + 8y = -5
-21 * (y-1) + 8y = -5
-21y + 21 + 8y = -5
21 + 5 = 21y - 8y
26 = 13y
y = 26/13
Therefore: y=2.
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If y=2,
x = 3(2-1) = 3
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Answers:
x=3 and y=2.
Answer:
5/6
Step-by-step explanation:
11/12 - 1/12 = 10/12
10 divided by 2 = 5
12 divided by 2 = 6
5/6
Answer:
dy/dx = (cos y + y sin x) / (cos x + x sin y)
Step-by-step explanation:
y cos x = x cos y
y (-sin x) + dy/dx cos x = x (-sin y dy/dx) + cos y
-y sin x + dy/dx cos x = -x sin y dy/dx + cos y
dy/dx (cos x + x sin y) = cos y + y sin x
dy/dx = (cos y + y sin x) / (cos x + x sin y)