![5ab^{3} +7-3a^{2} b ^{2} +a^{3}b+10](https://tex.z-dn.net/?f=5ab%5E%7B3%7D%20%2B7-3a%5E%7B2%7D%20b%20%5E%7B2%7D%20%2Ba%5E%7B3%7Db%2B10)
can be simplified to by adding the 7 and 10 to get
![a^{3}b-3a^{2} b ^{2}+ 5ab^{3} + 17](https://tex.z-dn.net/?f=a%5E%7B3%7Db-3a%5E%7B2%7D%20b%20%5E%7B2%7D%2B%205ab%5E%7B3%7D%20%2B%2017)
.
![5ab^{3}+3a^{2}b^{2}+a^{3}b -10](https://tex.z-dn.net/?f=5ab%5E%7B3%7D%2B3a%5E%7B2%7Db%5E%7B2%7D%2Ba%5E%7B3%7Db%20-10)
cannot be simplified any more by combining like terms.
By distributing the 2b into the parentheses, you can simplify the expression:
![5ab^{3}+2b(3ab^{2})+a^{3}b-10\\ 5ab^{3}+6ab^{3}+a^{3}b-10\\ 11ab^{3}+a^{3}b-10](https://tex.z-dn.net/?f=5ab%5E%7B3%7D%2B2b%283ab%5E%7B2%7D%29%2Ba%5E%7B3%7Db-10%5C%5C%0A5ab%5E%7B3%7D%2B6ab%5E%7B3%7D%2Ba%5E%7B3%7Db-10%5C%5C%0A11ab%5E%7B3%7D%2Ba%5E%7B3%7Db-10)
Here you can just add:
![5a^{3}b^{3}+3a^{2}b^{2}+a^{3}b^{3}-10ab\\ 6a^{3}b^{3}+3a^{2}b^{2}-10ab](https://tex.z-dn.net/?f=5a%5E%7B3%7Db%5E%7B3%7D%2B3a%5E%7B2%7Db%5E%7B2%7D%2Ba%5E%7B3%7Db%5E%7B3%7D-10ab%5C%5C%0A6a%5E%7B3%7Db%5E%7B3%7D%2B3a%5E%7B2%7Db%5E%7B2%7D-10ab)
Thus, the only expression that cannot simplify any more using adding like terms is the second,
![5ab^{3}+3a^{2}b^{2}+a^{3}b -10](https://tex.z-dn.net/?f=5ab%5E%7B3%7D%2B3a%5E%7B2%7Db%5E%7B2%7D%2Ba%5E%7B3%7Db%20-10)
.
Ok, so first this is a system. You have:
8x+12y=3200
And
X+y=300.
That means y= -X + 300.
If you replace "y" in the first equation with -X+300, it would be 8x+12(-X+300)=3200.
If you simplify that, X will be equal to 100. Now go back to the second equation, which is X+y= 300.
Now that you know X is 100, you can subtract it from 300 to find y.
So,
X=100 and y=200!
Hope this helps!!!
Answer:
No.
example: 123 ÷ 3 = 41 (2 digits)
425 ÷ 5 = 85 (2 digits)
but, if 428 ÷ 2 = 214, which gives us 3 digits but not 2
∴, the quotient doesn't always have the no. of digits when we divide 3 digit no. s by 1 digit no. s.
Answer:
2nd line
Step-by-step explanation:
The point is now at (3,0).