Answer: 90 + 4, third choice and 90 - -4, last choice
Her score is obviously higher when the mistake is fixed, adding the 4 points gives 94.
When you have something like this, all you need to do is substitute the values, the last is for what value of x
For the first one;
((x^2+1)+(x-2))(2)
(x^2+x-1)(2)
(2)^2+(2)-1
4+2-1
5
For the second one;
((x^2+1)-(x-2))(3)
(x^2-x+3)(3)
(3)^2-(3)+3
9-3+3
9
For the last one;
3(x^2+1)(7)+2(x-2)(3)
3((7)^2+7)+2((3)-2)
3(49+7)+2(3-2)
3(56)+2(1)
168+2
170
1) The expressions are not equivalent. When you expand and multiply 2(x + 3) it becomes 2x + 6. This is not equal to 3x + 5
2) They are equivalent. Again, expand and multiply the second expression. 2(3n + 4) becomes 6n + 8. This makes both sides equal.
3) They are equivalent. In the parentheses, you are adding 3 y's and a 2. This gives you 3y + 2. Now add the additional 3y that follows the closed parentheses. You'll have 6y + 2. Now both sides are equivalent.
Nadia = 3x than Jim
Paul = 25+ than Nadia
Jim = ?
Nadia and Paul = 85
85-25 = 60
(Subtract Nadia's and Paul's stickers all together with the stickers that Paul have more so both of the stickers are balance)
Nadia = 60/2 =30
Paul = 30+25 =55
Jim= 30/3=10
#Jim have 10 stickers
Answer:
there is no picture so I dont know what i have to do..
