All exercises involve the same concept, so I'll show you how to do the first, then you can apply the exact same logic to all the others.
The first thing you need to know is that, when a certain quantity multiplies a parenthesis, you can distribute that number to every element in the parenthesis. This means that
So, 
is multiplying the parenthesis involving 
and 
, and we distributed it: multiplies both and in the final result.
Secondly, you have to know how to recognize like terms, because they are the only terms you can sum. Two terms can be summed if they have the same literal expression. So, for example, you cannot sum 
, and neither 
exponents count.
But you can su, for example,
or
So, take for example exercise 9:
We distribute the 1.2 through the first parenthesis:
And you can distribute the negative sign through the second parenthesis (it counts as a -1 to distribute):
So, the expression becomes
Now sum like terms:
The little squares at corner-B and corner-E were drawn there
to show that those are right angles.
Answer:
B. (-3, 1), (6,2), (8,3), (6,4), (3, 5)
Step-by-step explanation:
This relation is not a function because the number 6 repeats twice.
Answer:
Sorrry man I need the points asap sorry again
A i think
Answer:
y = -2
Step-by-step explanation:
To find the equation of the tangent we apply implicit differentiation, and then we take apart dy/dx
The equation is
implicit differentiation give us
![\frac{d}{dx}[y^2(y^2-4)=x^2(x^2-5)]\\\\2y\frac{dy}{dx}(y^2-4)+y^2(2y\frac{dy}{dx})=2x(x^2-5)+x^2(2x)\\\\4y^3\frac{dy}{dx}-8y\frac{dy}{dx}=2x^3-10x+2x^3\\\\\frac{dy}{dx}=\frac{4x^3-10x}{4y^3-8y}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5By%5E2%28y%5E2-4%29%3Dx%5E2%28x%5E2-5%29%5D%5C%5C%5C%5C2y%5Cfrac%7Bdy%7D%7Bdx%7D%28y%5E2-4%29%2By%5E2%282y%5Cfrac%7Bdy%7D%7Bdx%7D%29%3D2x%28x%5E2-5%29%2Bx%5E2%282x%29%5C%5C%5C%5C4y%5E3%5Cfrac%7Bdy%7D%7Bdx%7D-8y%5Cfrac%7Bdy%7D%7Bdx%7D%3D2x%5E3-10x%2B2x%5E3%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cfrac%7B4x%5E3-10x%7D%7B4y%5E3-8y%7D)
But we know that
Hence, for the point (0,-2) and by replacing for dy/dx
Hence m=0, that is, the tangent line to the point is a horizontal line that cross the y axis for y=-2. The equation is:
y=(0)x+b = -2
HOPE THIS HELPS!!
