Answer:
Step-by-step explanation:
Note that the two equations are equivalent.
2y = 14-2x can be rearranged to y = -x+7, which is the same as the second equation. What you're looking is one line on top of the other.
Every shared point is a solution. Since there are infinitely many points on a line, and these lines share every point, there are infinitely many solutions.
The answer would be
x + 6.01
Hope this helps
Have a great day/night
well, we know the sine, and we also know that we're on the II Quadrant, let's recall that on the II Quadrant sine is positive whilst cosine is negative.
![\bf sin^2(\theta)+cos^2(\theta)=1~\hspace{10em} tan(\theta )=\cfrac{sin(\theta )}{cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ sin^2(a)+cos^2(a)=1\implies cos^2(a) = 1-sin^2(a) \\\\\\ cos^2(a) = 1-[sin(a)]^2\implies cos^2(a) = 1-\left( \cfrac{3}{4} \right)^2\implies cos^2(a) = 1-\cfrac{3^2}{4^2} \\\\\\ cos^2(a) = 1-\cfrac{9}{16}\implies cos^2(a) = \cfrac{7}{16}\implies cos(a)=\pm\sqrt{\cfrac{7}{16}}](https://tex.z-dn.net/?f=%5Cbf%20sin%5E2%28%5Ctheta%29%2Bcos%5E2%28%5Ctheta%29%3D1~%5Chspace%7B10em%7D%20tan%28%5Ctheta%20%29%3D%5Ccfrac%7Bsin%28%5Ctheta%20%29%7D%7Bcos%28%5Ctheta%20%29%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20sin%5E2%28a%29%2Bcos%5E2%28a%29%3D1%5Cimplies%20cos%5E2%28a%29%20%3D%201-sin%5E2%28a%29%20%5C%5C%5C%5C%5C%5C%20cos%5E2%28a%29%20%3D%201-%5Bsin%28a%29%5D%5E2%5Cimplies%20cos%5E2%28a%29%20%3D%201-%5Cleft%28%20%5Ccfrac%7B3%7D%7B4%7D%20%5Cright%29%5E2%5Cimplies%20cos%5E2%28a%29%20%3D%201-%5Ccfrac%7B3%5E2%7D%7B4%5E2%7D%20%5C%5C%5C%5C%5C%5C%20cos%5E2%28a%29%20%3D%201-%5Ccfrac%7B9%7D%7B16%7D%5Cimplies%20cos%5E2%28a%29%20%3D%20%5Ccfrac%7B7%7D%7B16%7D%5Cimplies%20cos%28a%29%3D%5Cpm%5Csqrt%7B%5Ccfrac%7B7%7D%7B16%7D%7D)
![\bf cos(a)=\pm\cfrac{\sqrt{7}}{\sqrt{16}}\implies cos(a)=\pm\cfrac{\sqrt{7}}{4}\implies \stackrel{\textit{on the II Quadrant}}{cos(a)=-\cfrac{\sqrt{7}}{4}}\\\\[-0.35em]~\dotfill\\\\tan(a)=\cfrac{sin(a)}{cos(a)}\implies tan(a)=\cfrac{~~\frac{3}{4}~~}{-\frac{\sqrt{7}}{4}}\implies tan(a)=\cfrac{3}{4}\cdot \cfrac{4}{-\sqrt{7}}\\\\\\tan(a)=-\cfrac{3}{\sqrt{7}}\implies \stackrel{\textit{rounded up}}{tan(a) = -1.13}](https://tex.z-dn.net/?f=%5Cbf%20cos%28a%29%3D%5Cpm%5Ccfrac%7B%5Csqrt%7B7%7D%7D%7B%5Csqrt%7B16%7D%7D%5Cimplies%20cos%28a%29%3D%5Cpm%5Ccfrac%7B%5Csqrt%7B7%7D%7D%7B4%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bon%20the%20II%20Quadrant%7D%7D%7Bcos%28a%29%3D-%5Ccfrac%7B%5Csqrt%7B7%7D%7D%7B4%7D%7D%5C%5C%5C%5C%5B-0.35em%5D~%5Cdotfill%5C%5C%5C%5Ctan%28a%29%3D%5Ccfrac%7Bsin%28a%29%7D%7Bcos%28a%29%7D%5Cimplies%20tan%28a%29%3D%5Ccfrac%7B~~%5Cfrac%7B3%7D%7B4%7D~~%7D%7B-%5Cfrac%7B%5Csqrt%7B7%7D%7D%7B4%7D%7D%5Cimplies%20tan%28a%29%3D%5Ccfrac%7B3%7D%7B4%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B-%5Csqrt%7B7%7D%7D%5C%5C%5C%5C%5C%5Ctan%28a%29%3D-%5Ccfrac%7B3%7D%7B%5Csqrt%7B7%7D%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7Btan%28a%29%20%3D%20-1.13%7D)
Answer:
units
Step-by-step explanation:
We are given;
- Coordinates of A (70, -80)
- Coordinates of B (-58, 48)
We are required to calculate the length of AB
- To determine the length of AB we are going to use the formula for getting magnitude;
- Given, coordinates (x₁, y₁) and (x₂, y₂), then
- Magnitude =√((x₂-x₁)²+(y₂-y₁)²)
Therefore;
Length = 
=
= 
= 
=
Therefore, the exact length of AB is units
The best fit curve for #1 is D. The easiest way to check these is by pluggin in to the equation and seeing if they come close. By doing just the first ordered pair along, it is apparent that only D will work.
With x = 1 input
A) -36
B) 24.2
C) 17.5
D) 11.58
#2 is also D. We can tell this because multiplying any of these options always results in a middle term between them. For instance, if you multiply out C, you will not only get x^4 and x^4, but you will also get terms such as 4x^2y^2 in the middle.
