The number in the ones place= X
The number in the tenths place= y
y+x=14.
y=14-x
Original placement of numbers=10y+x
New placement of numbers(after subtraction of 36)= 10x+y
10y+ x-36=10x+y
Substitute y with 14-x
So we would get
10(14-x)+ x-36=10x+(14-x)
=140-10x+x-36=10x+14-x
After transposing we get
140-14-36=10x-x+10x-x
=90=18x
X=90/18
x=5
Y=14-x
=14-5
=9
So the two digit number is
10y+x
=90+5
=95
It is answer D does that help?
Sounds like "formulas"! y = ax^2 + bx + c involves the variable x and the constants {a, b, c}. The "2" indicates "squaring function."
Answer:
Let the integers are p and q.
In case both integers are positive or negative, you add the numbers up and apply their sign to the sum.
<u>Examples:</u>
- 10 + 88 = 98
- -20 + (-15) = -35
In case one of the integers is negative and one positive.
Use absolute value in this case.
Subtract the numbers and apply the sign of the number with the greater absolute value.
<u>Examples:</u>
- - 10 + 15 = |15 - 10| = |5| = 5
- - 15 + 10 = - |15 - 10| = - |5| = -5
answer - true
all rectangles have four sides, and a shape is a quadrilateral if it has four sides
therefore, all rectangles are quadrilaterals
