Okay so combine like terms
10x-4x= 6x
2-8= -6
6x=-6
x= -1
Now to check:
10(-1)-2=4(-1)-8
-10-2=-4-8
-12=-12
<span>the limit as x approaches -3 of [g(x)-g(-3)]over(x+3) is the same as the derivative, or slope, of g(x) at the point x=-3, or g'(-3).
Since you are given the equation of the tangent line, the answer is just the slope of that line.
</span><span>2y+3=-(2/3)(x-3)
</span><span>6y+9=-2(x-3)
6y+9=-2x+6
6y=-2x-3
y= (-2x-3)/6
slope is -2/6 = </span>
Answer:

Step-by-step explanation:
Given
Negative integer J
Required
Represent as an inequality of its inverse
The question didn't state if it's additive inverse or multiplicative inverse;
<em>Since the question has to do with negation, I'll assume it's an additive inverse</em>
<em></em>
The inverse of -J is +J
To represent as an inequality (less than or equal), we have:

Solving further, it gives

Answer:
48
Step-by-step explanation:
a=2
b=3
3a^2+4b^2
3(2)^2+4(3)^2
3(2)(2)+4(3)(3)
3(4)+4(9)
12+36
48
Hope this helps ;) ❤❤❤