Let the first number = x
The second number would be 5x
The third number would be 100x
Add:
X + 5x + 100+x = 296
Simplify:
7x + 100 = 296
subtract 100 from both sides:
7x = 196
Divide both sides by 7
X = 196 / 7
X = 28
First number is 28
Second number is 28 x 5 = 140
Third number = 128
If you do 5/5-4/5 that will equal 1/5.
Not easy to graph but
multily first equaiton by -2 remember to flip sign
-2x+4y>-6
add to other equaiton
2x-2x+y+4y>-6+8
5y>2
divide by 5
y>2/5
subtitute
2x+2/5>8
subtract 2/5 from both sdies
2x>7 3/5
divid eby 2
x>38/10=19/5=3 4/5
x>3 4/5
y>2/5
just shade the area in that zone, not including point (0,0)
![\\ \tt\longmapsto 2n=3k](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Clongmapsto%202n%3D3k)
![\\ \tt\longmapsto n\propto k](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Clongmapsto%20n%5Cpropto%20k)
- Both n and k are positive i.e n,k>0
So the domain of n is
![\\ \tt\longmapsto n\in [0,100]](https://tex.z-dn.net/?f=%5C%5C%20%5Ctt%5Clongmapsto%20n%5Cin%20%5B0%2C100%5D)
Now
- N has total 100+1=101 integer values
- K will also have 100+1=101values .
So total solution sets are 101 .
Answer:
i think the answer is 8
Step-by-step explanation:
"three less than the quotient of 6 and a number (n=3) increased by nine"
(6÷3)-3+9
the quotient of 6 and 3 (6÷3) is 2
three less than 2: 2-3 which equals -1
increased by nine: -1+9 (which is also the same as 9-1) is 8
please don't hesitate to ask for clarification! hope i helped!