Volume of the cone:
V = 3.14 x 3^2 x 9/3 = 84.78
Volume of half sphere on top of cone:
V = (4/3 x 3.14 x 3^3 )/2 = 56.52
Total volume = 56.52 + 84.78 = 141.3 cm^3
24 because 12 + 12 = 24 and and CD = DB
CB = CD + CB = 24 the triange ADC and ADB are equal
Answer:
i need more info
Step-by-step explanation:
Problem One
Remark
If she doesn't mind having I kg left over, the minimum number would be 3 five kg boxes. If on the other hand, she must have exactly 14 kg then the minimum number is 6.
She needs 2 five kg boxes and 4 one kg boxes. <<<< Answer
Problem Two
There is a method of solving this that is called dimensional analysis. It is what should be used here. I'll do it at the end of the problem. In the meantime, you have to do it a slightly longer way.
1 portion = 100 grams.
x portions = 1kg which is 1000 grams.
x portions = 1000 grams.
Set up a proportion to find the number of servings in 1 kg
1 portion/x = 100 grams/1000 grams Cross multiply
1 * 1000 = 100 * x Divide by 100
1000/100 = x
x = 10 servings in 1 kg.
So each kg produces 10 portions
1 kg / 10 portions = 20 kg / x portions Cross multiply
x * 1 = 10 * 20
x = 200 portions <<<<< Answer
Dimensional Analysis
[1batch]*[1 portion/100g][1000g/kg][20kg/batch] the units cancel
1000 * 20 / 100 only the portions are left over.
200 portions is the answer.
Problem Three
1 kg = 1000 grams
x kg = 5000 grams Cross multiply
1*5000 = 1000 x
x = 5 kg
1 parcel weighs 5 kg
x = 15 kg
15 kg = 5 x
x = 15/5
x = 3
So he can carry 3 parcels per trip.
Since there are 5 such parcels, he will have to make 2 trips. The second one will not be a full load.
First Trip = 3 parcels
Second Trip = 2 parcels. <<<<Answer
