1. Find three different solutions of the each of the following equations.i) 3x + 4y = 7ii) y=6xiv)

(V+1)2 is that right

(x - h)^2 + (y - v)^2 = r^2.

The answer would be C.

Hope this helps! :)

 The required value of the given expression is -204.

Step-by-step explanation:  We are given the following two vectors :

[tex]u=-8i+3j,\\\\v=6i-j.[/tex]

We are to find the value of the following :

[tex](u+v)^2-(u-v)^2.[/tex]

We will be using the following identities :

[tex]i.j=j.i=0,\\\\i^2=j^2=1.[/tex]

Therefore, the value of the given expression can be evaluated as follows :

[tex](u+v)^2-(u-v)^2\\\\=(u^2+2uv+v^2)-(u^2-2uv+v^2)\\\\=4uv\\\\=4(-8i+3j)(6i-j)\\\\=4(-48i^2+8ij+18ji-3j^2)\\\\=4(-48\times1-3\times1)\\\\=4(-48-3)\\\\=-204.[/tex]

Thus, the required value of the given expression is -204.

[tex]c) (x-h)^{2}+(y-v)^{2}=r^{2}[/tex]

Step-by-step explanation:

Hi There,

1. The Circle formula in its standard form is given by:

[tex](x-a)^{2}+(y-b)^{2}=r^{2}\Rightarrow C=(a,b)[/tex]

2) This coordinates a, b  are the Center coordinates of the Center, a point distant from the circumference by the radius.

3) Because we can find derive this formula from that. (Check the graph below)

There's a point P(x,y) whose distance to C(h,v) is the radius, we need to calculate it numerically:

[tex]r=\sqrt{(x-h)^{2}+(y-v)^{2}} \:or\:r^{2}=(x-h)^{2}+(y-v)^{2}\\r=\sqrt{x^{2}-2hx+h^{2}+y^{2}+2vy+v^{2}}\\r=\sqrt{x^{2}+h^{2}+y^{2}+v^{2}}\\(r)^{2}=(\sqrt{x^{2}+h^{2}+y^{2}+v^{2}})^{2}\\r^{2}=x^{2}+h^{2}+y^{2}+v^{2}\\r^{2}=x^{2}+y^{2}+h^{2}+v^{2}\\\\[/tex]

4) Hence, as the Center has its coordinates C(h,v) then its circle formula is:

[tex]c) (x-h)^{2}+(y-v)^{2}=r^{2}[/tex]

answer for the first one

The answer is (v+1)^2

(side note: when writing exponents use "^" to indicate the following number is an exponent. ex. 3 squared would be 3^2)