Answer:
Expert Verified
Step-by-step explanation:
the diameter is 57.2 in, then the radius is half that, or 28.6 in. The volume of a sphere is V = (4/3)πr³, and so the volume of this hemisphere is (2/3)πr³
Answer:
x = 6
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define Equation</u>
3 - 2x = -1.5x
<u>Step 2: Solve for </u><em><u>x</u></em>
- Add 2x on both sides: 3 = 0.5x
- Divide 0.5 on both sides: 6 = x
- Rewrite: x = 6
<u>Step 3: Check</u>
<em>Plug in x into the original equation to verify it's a solution.</em>
- Substitute in<em> x</em>: 3 - 2(6) = -1.5(6)
- Multiply: 3 - 12 = -9
- Subtract: -9 = -9
Here we see that -9 does indeed equal -9.
∴ x = 6 is the solution to the equation.
Answer:
√20
Or
2√5
Step-by-step explanation:
It is very easy just substitute
It will be:
Let the one type of the bread be bread A
The second type of the bread be bread B
Let the flour be 'f' and the butter be 'b'
We need 150f + 50b for bread A and 75f + 75b for bread B
We can compare the amount of flour and bread needed for each bread and write them as ratio
FLOUR
Bread A : Bread B
150 : 75
2 : 1
We have a total of 2250gr of flour, and this amount is to be divided into the ratio of 2 parts : 1 part. There is a total of 3 parts.
2250 ÷ 3 = 750 gr for one part then multiply back into the ratio to get
Bread A : Bread B = (2×750) : (1×750) = 1500 : 750
BUTTER
Bread A : Bread B = 50 : 75 = 2 : 3
The amount of butter available, 1250 gr is to be divided into 2 parts : 3 parts.
There are 5 parts in total
1250 ÷ 5 = 250 gr for one part, then multiply this back into the ratio
Bread A: Bread B = (2×250) : (3×250) = 500 : 750
Hence, for bread A we need 1500 gr of flour and 500 gr of butter, and for bread B, we need 750 gr of flour and 750 gr of butter.
Answer:
a: given
b: definition of congruent
c: definition of bisect
Step-by-step explanation:
just took the assignment
