Let's call the first unknown integer x and the second one y.
"Four times an unknown integer"
"Multiplied by three times the unknown integer plus a different unknown integer"
- 4x is multiplied by 3x+y, so 4x(3x+y)=12x²+4xy
"Equals 100."
And so the answer is C: 12x²+4xy-100=0
Answer:
The larger number is -6, the smaller number is -15
Step-by-step explanation:
We have two numbers, a and b.
We know that one number is larger than another by 9.
Then we can write:
a = b + 9
then a is larger than b by 9 units.
If the greater number is increased by 10 (a + 10) and the lesser number is tripled (3*b), the sum of the two would be -41:
(a + 10) + 3*b = -41
So we got two equations:
a = b + 9
(a + 10) + 3*b = -41
This is a system of equations.
One way to solve this is first isolate one variable in one of the two equations:
But we can see that the variable "a" is already isolated in the first equation, so we have:
a = b + 9
now we can replace that in the other equation:
(a + 10) + 3*b = -41
(b + 9) + 10 + 3*b = -41
now we can solve this for b.
9 + b + 10 + 3b = -41
(9 + 10) + (3b + b) = -41
19 + 4b = -41
4b = -41 -19 = -60
b = -60/4 = -15
b = -15
then:
a = b + 9
a = -15 + 9 = -6
a = -6
Answer:
xx(0.65) + xx(0.08)
Step-by-step explanation:
First step is to find the discounted price. Finding 35% less than the original xx is the same as finding 65% of it, so we multiply by 0.65. Once this is completed we need to find 8% more than that. So we add 8% of xx, which is the same as multiplying by 0.08. Hope this helps :)
Answer:
1. 125
2. 25
3. 40
4. 39
First question:
(15-10)^2 + (15-5)^2
parentheses first to get 5^2 + (15-5)^2
exponents next to get 25 + (15-5)^2
parentheses again 25 + 10^2
exponents again 25 + 100 = 125
Second question:
simplify the exponents (3 *3) + (4 * 4) + (2 * 2) to get 9 + 16 + 4 = 29
Third question:
simplify in the parentheses 6 1/7 - 1/7 = 6
divide 60 by 6 to get 10
multiply 10 by 4 to get 40!
Fourth question:
simplify the fractions to get 1/3 + 8/3 = 3
multiply the 3 by 13 to get 39
4 5/12 you just add the fractions
