Well obviously quarters equal 25 cents. So you have to divide the 8.85 by 25 first. That means he could have 35 quarters. That least the rest as being dimes. So find the remaining amount of the 8.85. If you have 35 quarters that's $8.75. And 8.85 minus 8.75 is 10 cents. which would be 2 nickels. That only adds up 37 coins. So you break down one of the 25. That would mean 5 more coins could be nickels. Add 37 and 5 to get 42. Do it again. 25 cents in nickels would be 5 more coins. would be 46. that would be missing 2 coins. But don't forget you're also subtracting from the 35 quarters. Now you would technically have 12 nickels and 33 quarters. So you can do it one more time. 32 quarters means there is $8 in quarters at this point.
Now add you have 12 nickels, and you made 5 more. That's 17 nickels. That equals 85 cents. So add 17 and 32 to be sure you have 49 coins, which you do.
So 17 nickels, 32 quarters to equal 49 coins and $8.85
(sorry some of it got mixed up above because I was forgetting to subtract the quarters from the total as I changed them to nickels, so pay attention to the end)
Answer:
1) 5f
2) 45gm-7g
Step-by-step explanation:
hope this helped!
The correct answer is
the flagpole is <span>
33 feet high</span>.
Explanation:
Please refer to the attached picture.
We know:
CD = 40 feet
AC = 5 feet
∠BDC = α = 35°
Using trigonometry, we know that the definition of the tangent of an angle is the ratio between the opposite side and the adjacent side, therefore:
tan α = BC / CD
Solving for BC:
BC = CD · <span>tan α
= 40 </span>· tan (35)
= 28 feet
In order to find the height of the flagpole, we need to add the distance of the clinometer from the ground:
AB = BC + AC
= 28 + 5
= 33
Hence, the flagpole is
33 feet high.
Answer:
Step-by-step explanation:
Volume of the butter in the cardboard = 31.25 cubic inches
Length of the Molded Square Prism =5 Inches
Let the length of the square cross sectional area = x
Volume of the Prism
Since l=5 Inches

Next, we determine the surface area of the New Cardboard box to be made.
Length of the Cardboard Box=5 Inches
Length of the square cross sectional area = 2.5 Inches
Total Surface Area

Therefore,
of cardboard is needed,