Answer:
<em>Y alone can do the piece of work in 30 days.</em>
Step-by-step explanation:
<u>Proportions</u>
Let's make:
N = days for Y to finish the work alone
Since X alone can do it in 15 days, each day he can do a proportion of 1/15 of the piece of work.
Since Y alone can (possibly) do it in N days, each day he can do 1/N parts of the work.
Together, they do
parts of the work per day. We know they can finish it in 10 days, thus:
Rearranging:
The LCM of 10 and 15 is 30, thus operating:
Solving for N:
N = 30
Y alone can do the piece of work in 30 days.
Okay, you will need to use the law of cosines for this problem.
The Law of Cosines states (in this case): a^2 = b^2 + c^2 - 2 * b * c * cos A, where "a" is the side opposite angle A (7 inches), and b and c are the other two sides.
Plug the numbers in and you get: 7^2 = 5^2 + 9^2 - 2 * 5 * 9 * cos A, or:
49 = 25 + 81 - 90 * cos A.
Subtract (25 + 81) from both sides to get:
-57 = -90 * cos A.
Divide by -90 on both sides:
cos A = 19/30
To find A, you do the inverse trigonometric function to get:
cos^-1 of (19/30) = A.
I don't have a calculator that can do this right now, but if you plug the left side of the above equation into it (make sure it is in degrees, not radians), you should get A.
Answer:
elasticity supply of dog food = 2.61
elasticity supply of cat food = 1.71
Step-by-step explanation:
The midpoint formula for elasticity is:
![Elasticity = \frac{(Q2-Q1)/[(Q2+Q1)/2]}{(P2-P1)/[(P2+P1)/2]}](https://tex.z-dn.net/?f=Elasticity%20%3D%20%5Cfrac%7B%28Q2-Q1%29%2F%5B%28Q2%2BQ1%29%2F2%5D%7D%7B%28P2-P1%29%2F%5B%28P2%2BP1%29%2F2%5D%7D)
Point 1: Q = 39.0 and P = 5.50
Point 2: Q = 101.0 and P = 7.75
![Elasticity\ supply\ of\ dog\ food = \frac{(101.0-39.0)/[101.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=2.61](https://tex.z-dn.net/?f=Elasticity%5C%20supply%5C%20of%5C%20dog%5C%20food%20%3D%20%5Cfrac%7B%28101.0-39.0%29%2F%5B101.0%2B39.0%29%2F2%5D%7D%7B%287.75-5.50%29%2F%5B%287.75%2B5.50%29%2F2%5D%7D%3D2.61)
Doing the same for the cat food:
![Elasticity\ supply\ of\ cat\ food = \frac{(71.0-39.0)/[71.0+39.0)/2]}{(7.75-5.50)/[(7.75+5.50)/2]}=1.71](https://tex.z-dn.net/?f=Elasticity%5C%20supply%5C%20of%5C%20cat%5C%20food%20%3D%20%5Cfrac%7B%2871.0-39.0%29%2F%5B71.0%2B39.0%29%2F2%5D%7D%7B%287.75-5.50%29%2F%5B%287.75%2B5.50%29%2F2%5D%7D%3D1.71)
It's simple .....
just subtract R and C.....
so 50x+10-10x-275
= 40x-265
so the answer is B
0.5, 0.05 is 1/10 the value of 0.5
