When you are looking at a graph, a minimum point would be where the curve is decreasing, then begins to increase. Right at the point where it switches, the slope is a horizontal line, or 0. We can take the derivative is f(x), then look for all the x values where the slope (which is equal to the first derivative) is equal to zero.
f'(x) = 2 * -4sin(2x - pi)
The 2 comes from the derivative of the inside, 2x-pi.
So now set the derivative equal to 0.
-8sin(2x-pi) = 0
We can drop the -8 by dividing both sides by -8.
sin(2x-pi) = 0
This can be rewritten as arcsin(0) = 2x-pi
So when theta equals 0, what is the value of sin(theta)? At an angle of 0, there is just a horizontal line pointing to the right on the unit circle with length of 1. Sine is y/h, but there is no y value so it is just 0. If arcsin(0) = 0, we can now set 2x-pi = 0
2x = pi
x = pi/2
This is a critical number. To find the minimum value between 0 and pi, we need to find the y values for the endpoints and the critical number.
f(0) = -4
f(pi/2) = 4
f(pi) = -4
So the minimum points are at x=0 and x=pi
Answer:
x≤50
Step-by-step explanation:
100-3x≥-50
subtracting 100 at both side on inequality
-100+100-3x≥--50-100
-3x≥-150
Dividing both sides by 3
-3x/3≥-150/3
-x≥--50
Adding 50 on both sides
-x+50≥-50+50
-x+50≥0
Adding x on both sides
-x+x+50≥0+x
50≥x
so,
x≤50
Answer:
-1
Step-by-step explanation:
Tip: Remember to always start from the inside, which would be g(3), in this case.
The first step in solving this problem is to solve for g(3).
To accomplish this, you must substitute 3 for x into the given equation g(x) = x^2 - 10
- g(3) = 3^2 - 10
- g(3) = 9 - 10
- g(3) = -1
The next step is to substitute the answer of g(3), -1, for x in the given equation f(x) = 2x + 1.
Because the equation is asking for f[g(3)], it becomes f(-1) because g(3) = -1.
- f(-1) = 2(-1) + 1
- f(-1) = -2 + 1
- f(-1) = -1
Therefore, f[g(3)], or f(-1), equals -1
Answer:
The area of the shaded region is 
Step-by-step explanation:
we know that
The area of the shaded region is equal to the area of the larger circle minus the area of the square plus the area of the smaller circle
Step 1
<em>Find the area of the larger circle</em>
The area of the circle is equal to
we have
substitute in the formula
step 2
<em>Find the length of each side of square BCDE</em>
we have that
The diagonal DB is equal to
Let
x------> the length side of the square BCDE
Applying the Pythagoras Theorem
step 3
<em>Find the area of the square BCDE</em>
The area of the square is
step 4
<em>Find the area of the smaller circle</em>
The area of the circle is equal to
we have
substitute in the formula
step 5
Find the area of the shaded region
Answer:
- <u><em>Rounding to nearest tenth of centimeter, the ball bounces 192.1 cm high on the 5th bounce.</em></u>
<u><em></em></u>
Explanation:
The <em>ball is dropped from a height of 900 centimeters</em>.
Since the heights form a <em>geometric sequence,</em> you can find a common ratio between consecutive terms. This is:
- Height bounce 2 / height bounce 1 = 560 / 800 = 0.7
- Height bound 3 / height bounce 2 = 392 / 560 = 0.7
Hence, the ratio of the geometric sequence is 0.7, and taking bounce 1 as the start of the sequence, the general term of the sequence is:
With that formula you can find any term:
Rounding to <em>nearest tenth of centimeter</em>, the ball bounces 192.1 cm high on the 5th bounce.
