So,
We are trying to figure out when Grandpa Lopez's age was twice that of Dad.
Let x represent the number of years before/after when G. Lopez's age was twice that of Dad.
66 + x = 2(37 + x)
Distribute.
66 + x = 74 + 2x
Subtract x from both sides.
66 = 74 + x
Subtract 74 from both sides.
-8 = x
So 8 years ago, G. Lopez was twice as old as Dad. Let's check that.
66 - 8 = 58
37 = 8 = 29
29 * 2 = 58
58 = 58
It checks.
Answer:
<h2>there are 22 dimes</h2><h2>there are 44 nickles</h2><h2 />
Step-by-step explanation:
D be dimes and N be nickles
2 nickles as dime
nickle = 5 cents and dime =10 cents
N=2D
0.10 D + 0.05 (N)=4.4 substitute N=2D
0.10 D +0.05(2D)=4.4
0.10D+0.10D=4.4
0.20 D=4.4
D=4.4/0.2 = 22
<h2>there are 22 dimes</h2>
N=2D
N=2(22)
N=44
<h2>there are 44 nickles</h2>
check : 0.10(22)+0.05(44)= 4.4 (correct)
Since 4pi/3 = 3pi/3 + pi/3 = pi + pi/3, that means it goes past the angle of pi (the negative x-axis), and an additional pi/3 radians. So this gives you a diagonal line that passes through the origin, from the third quadrant through to the first quadrant, and makes an angle of 60 degrees with the negative x-axis.
85 + 91 + x ÷3 ≥94
You didn't give any choices but this represents the problem.
V= piR^2(h)
6283=pi(10^2)(h)
6283=pi(100)h
height = 6283/(pi(100))
then round to the nearest inch
