Answer:
The answer is 8 quarters, 16 nickles, and 32 dimes
Step-by-step explanation:
- First, the equation would be (2/4d) x 5 +d x 10 +1/4d x 25= 600 (Don't put 6.00, that's dollars. We put 6 dollars into cents to make it easier.)
- Then we slowly start to solve: 10/4d + 10d + 25/4d = 600
- 10d+40d+25d/4 = 600 (Put all of that over 4, not first 25d)
- 75d= 600 x 4
- d=600 x 4/75 (same here put everything over 75)
- d=32
- So then n= 2 x 32/4 = 16
- Then n would be equal to : 1/4d= 1/4 x 32 = 8
- So your answer would be 8 quarters, 16 nickles, and 32 dimes.
Hope this helped :)
The answer is c first 2×6= 12 then 12-5=7
then do -7×-2=14 the minus signs cancel
finaly we have two bracket so we multiply both number (7)×(14)=98
A number that is greater than 714,587 is 1,000,000
Check the picture below.
a)
so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.
the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".
let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.
and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.


b)
we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.

Answer:
A) 2 miles, 5 miles, 8 miles
Step-by-step explanation:
Let's get some details cleared.
We're interested in finding how many miles he walked at the end of each hour in a period of 3 hours.
We know what is, so we can use that as our starting point.
Your answer is A.