So,
Proportion basically means, that something is proportional
Example: A rope's length and weight are in proportion.
It's in proportion because the longer the rope, the more the weight
so, let's say
A popcorn machine costs 20 an hour
And in order to rent one, you have to pay $32 dollars
So you pay 32 for the machine, then 20 for every hour you use it
the number of hours used WOULD be proportional to the cost because
the more hours used, the more the cost
your final answer: YES it is proportional
Answer:
f
Step-by-step explanation:
because you can substitute f by q-7
Answer:
A) see attachment
B)a, d
Step-by-step explanation:
A) The attachment shows the stem and leaf diagram
B) The data is strongly on lower side that is from 20% to 60% and is therefore skewed to left.
Answer:
75 degrees
Step-by-step explanation:
those two angles = 180
180-105 =75
Answer:
a) -13.9 ft/s
b) 13.9 ft/s
Step-by-step explanation:
a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:
(1)
![\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28D%5E%7B2%7D%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%2890%20-%20x%29%5E%7B2%7D%20%2B%2090%5E%7B2%7D%29%7D%7Bdt%7D)
(2)
Since:
![D = \sqrt{(90 -x)^{2} + 90^{2}}](https://tex.z-dn.net/?f=%20D%20%3D%20%5Csqrt%7B%2890%20-x%29%5E%7B2%7D%20%2B%2090%5E%7B2%7D%7D%20)
When x = 45 (the batter is halfway to first base), D is:
![D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62](https://tex.z-dn.net/?f=%20D%20%3D%20%5Csqrt%7B%2890%20-%2045%29%5E%7B2%7D%20%2B%2090%5E%7B2%7D%7D%20%3D%20100.%2062%20)
Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:
![\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%28D%29%7D%7Bdt%7D%20%3D%20-%5Cfrac%7B%2890%20-%2045%29%2A31%7D%7B100.62%7D%20%3D%20-13.9%20ft%2Fs)
Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.
b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:
![\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%28D%5E%7B2%7D%29%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%2890%5E%7B2%7D%20%2B%20x%5E%7B2%7D%29%7D%7Bdt%7D%20)
(3)
We have that D is:
![D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63](https://tex.z-dn.net/?f=D%20%3D%20%5Csqrt%7Bx%5E%7B2%7D%20%2B%2090%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B%2845%29%5E%7B2%7D%20%2B%2090%5E%7B2%7D%7D%20%3D%20100.63)
By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:
![\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s](https://tex.z-dn.net/?f=%5Cfrac%7BdD%7D%7Bdt%7D%20%3D%20%5Cfrac%7B45%2A31%7D%7B100.63%7D%20%3D%2013.9%20ft%2Fs)
Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.
I hope it helps you!