Step-by-step explanation:
For the triangle on the bottom right the missing angle is
180- (74+50)= 56°
For the triangle on the bottom left the missing angle is
180- (45+80)= 55°
For the triangle in the middle the missing angle is
180- (54+51)= 75°
For the triangle on top the missing angle is
180- (80+54)= 46°
180- (74+51)= 55°
180- (46+55)= 79°
First, we need to transform the equation into its standard form (x - h)²=4p(y - k).
Using completing the square method:
y = -14x² - 2x - 2
y = -14(x² + 2x/14) - 2
y = -14(x² + 2x/14 + (2/28)²) -2 + (2/28)²
y = -14(x + 1/14)² - 391/196
-1/14(y + 391/196) = (x + 1/14)²
This is a vertical parabola and its focus <span>(h, k + p) is (-1/14, -391/196 + 1/56) = (-1/14, -775/392).
Or (-0.071,-1.977).</span>
• The ball is at the same height as the building between 8 and 10 seconds after it is thrown. TRUE - the height is zero somewhere in that interval, hence the ball is the same height from which it was thrown, the height of the roof of the building.
• The height of the ball decreases and then increases. FALSE - at t=2, the height is greater than at t=0.
• The ball reaches its maximum height about 4 seconds after it is thrown. TRUE - the largest number in the table corresponds to t=4.
• The ball hits the ground between 8 and 10 seconds after it is thrown. FALSE - see statement 1.
• The height of the building is 81.6 meters. FALSE - the maximum height above the building is 81.6 meters. Since the ball continues its travel to a distance 225.6 meters below the roof of the building, the building is at least that high.
1. TRUE
2. False
3. TRUE
4. False
5. False
I think it's higher because as far as the integer gets, the higher its absolute value gets. So, put higher in the blank.
<span>The graph you plotted is the graph of f ' (x) and NOT f(x) itself. </span>
Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get
f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24
So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5
<span>f ' (x) = (x-4)(6-2x)
</span><span>f ' (3.5) = (3.5-4)(6-2*3.5)
</span>f ' (3.5) = 0.5
The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.
So the final answer is C) 3
