Answer:
The first set is a set of linear equations.
The way to figure this out is pretty easy. If you want to see it visually, go search up desmos graphing calculator and put in these equations.
A linear equation is a function that has a constant slope, meaning that the rate it increases or decreases will never change. The first one is a set of linear equations because it is 2 equations with constant slopes, meaning that the slopes will never change no matter what y and x are.
The second set is not, because while the first equation is linear, the second is an inequality. While it is a straight line, it doesn't count as a linear equation.
The third set, both equations have exponents on the x, which means that the slope will change depending on x. This means that both of these are not linear equations.
The only set that is a linear set is the one that has only linear equations.
we'll start off by grouping some
so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?
well, let's recall that a perfect square trinomial is
so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.
so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²
![\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29%3D6%5Cimplies%20%5Cleft%28%20x%5E2%20-x%20%2B%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%20%5Cright%29-%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%5D%5E2%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D6%2B%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Cleft%28x-%5Ccfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%3D%5Ccfrac%7B25%7D%7B4%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Ccfrac%7B25%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B%5Csqrt%7B25%7D%7D%7B%5Csqrt%7B4%7D%7D%5Cimplies%20x-%5Ccfrac%7B1%7D%7B2%7D%3D%5Ccfrac%7B5%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B5%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B2%7D%5Cimplies%20x%3D%5Ccfrac%7B6%7D%7B2%7D%5Cimplies%20%5Cboxed%7Bx%3D3%7D)
<h2>
Area of rectangle is 10L-L²</h2>
Step-by-step explanation:
Let L be the length and W be the width.
Perimeter of rectangle = 2 ( L + W )
That is
2 ( L + W ) = 20
L + W = 10
W = 10 - L
We need to express the area of the rectangle as a function of the length of one of its sides.
Area = Length x Width
A = LW
A = L(10 - L) = 10L-L²
Area of rectangle is 10L-L²
S=4π<span>r<span>2 - this is the equation for the surface area of a sphere. We know that the radius (r) is 3.5 so here's how to solve this:
S = 4</span></span><span>π r2
S = 4</span><span>π (3.5)2
S = 4</span><span>π 12.25
S = 49</span><span>π
The surface area of a sphere with radius of 3.5 is 49</span><span>π.</span>
5.19 / 16.8 = <span>0.30892857142</span>
