Answer:
Modays
Step-by-step explanation:
They suck.
Answer:
Step-by-step explanation:
hello :
factor the following polynomial is : (a+b+c) so :
3a(a+b+c)-2b(a+b+c)-c(a+b+c) = (a+b+c)(3a-2b-c)
Type A: $20.20 ÷ 10 gal = $2.02 per gal
Type B: $$26.04 ÷ 12 gal = $2.17 per gal
Type C: $ 28.14 ÷ 14 gal = $2.01 per gal
Type D: $30.45 ÷ 15 gal = $2.03 per gal
Type B costs less per gallon.
Hope this helps!!
1/4a+1/3=1/3a+5/4
1/4a-1/3a+1/3=1/3a-1/3a+5/4
1/4a-1/3a+1/3=5/4
1/4a-1/3a+1/3-1/3=5/4-1/3
1/4a-1/3a=5/4-1/3
Figure the common denominator for 1/4a,1/3a, 5/4 and -1/3 which is 12
Multiply by 3 for 1/4a. (1)(3)/4(3a)=3/12a
Multiply by 4 for 1/3a. (1)(4)(3)(4a)=4/12a
Multiply by 3 for 5/4. (5)(3)/4(3)=15/12
Multiply by 4 for -1/3. (-1)(4)(3)(4)=-4/12
3/12a-4/12a=15/12-4/12
-1/12a=11/12
Multiply by -12/1 for -1/12a and 11/12
-1/12(-12/1)a=11/12(-12/1)
Cross out 12 and -12 , divide by 12 and then becomes -11.
Cross out -12 and -12 , divide by 12 and then becomes 1.
Cross out -1 and 1, divide by 1 and then becomes 1.
1*1*a=-11
a=-11
Answer: a=-11
The correct answer is A. 1.5 cups
Explanation:
The total of polishing wooden furniture is equivalent to the total of olive oil plus the total of lemon juice. Additionally, in this case, the result should be given in cups. First, let's consider the values given.
50% of a cup of lemon juice + 25% of a quart with olive oil
The first quantity 50% of a cup of lemon indicates half cup of lemon was added, which is equivalent to 0.5 cups as 1 cup = 100% of a cup and 50% is the half.
Now, the second quantity is provided in quarts, which is not the same as a cup. Indeed 1 quart is equivalent to 4 cups, which represents 100%. This means 100% = 4 cups. Now, to calculate the 25% follow this procedure.
1. Write the values
4 = 100 (total percentage)
x = 25 (percentage to be calculated
2. Solve this by using cross multiplication
x100 = 4 × 25
x100 = 100
x = 100 ÷ 100
x = 1 (cups in 25%)
This means 1 cup of olive oil was added and this plus 0.5 cup of lemon juice is equivalent to 1.5 cups of solution.