<span>(11*.037+1)*490.28= 689.82
Hope I helped :)</span>
Answer:
7 and 11
Step-by-step explanation:
7+11=18
11-7=4
200,000 is ten times greater then 20,000
Answer:
x+8 is the oblique asymptote.
Step-by-step explanation:
we have to find the oblique asymptote for the given function :
Oblique asymptote case arises when degree of numerator is greater than denominator
We divide the numerator by denominator to find the oblique asymptote.
You can see the long division in the attachement.
The quotient will give an equation which will be our required oblique asymptote.
Hence, x+8 is the oblique asymptote.
Answer:
the consecutive integars is 11,12,13,14,15
Step-by-step explanation:
- let the first integar be described as y
- since they are consecutive ( following each other)
- the second will y + 1
- the third is y + 2
- the fourth is y + 3
- the fifth y + 4
then it says the sum of the first and 4 times the third = 60 less than 3 times the sum of the second, fourth and fifth
60 less than 3 times the sum of the second, fourth and fifth means that 3 times the sum of the second, fourth and fifth minus 60
y+ 4(y+2) = 3{ (y +1) +(y+3) + (y+4)} -60
y + 4y + 8 =3[ y +1 + y + 3 +y+4} - 60
5y +8=3(y+y+y+1+3+4)-60
5y+8=3(3y+8)-60
- simplify and open the brackets
5y+8=9y+24-60
5y+8=9y-36
- subtract 8 from both sides
5y+8-8=9y-36-8
5y=9y-44
- subtact 9y from both sides
5y-9y=9y-9y-44
-4y=-44
- divide -4 from both sides
-4y/-4=-44/-4 ( minus divided by minus is plus
y = 11
- the first integar is y = 11
- the second will y + 1 = 11+1 = 12
- the third is y + 2 = 11 +2 = 13
- the fourth is y + 3 = 11 + 3 = 14
- the fifth y + 4 = 11 + 4 = 15
- the consecutive integars is 11,12,13,14,15