To solve this, we must isolate x (have it by itself).
4/5x = 56
Multiply each side by 5x.
4/5x(5x) = 56(5x)
Divide each side by 280.
4/280 = 280x/280
Solve.
0.014 = x
0.014 can also be written as 1/70
:)
![\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad cos(\theta)=\cfrac{adjacent}{hypotenuse} \quad % tangent tan(\theta)=\cfrac{opposite}{adjacent}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Bhypotenuse%7D%0A%5Cqquad%0Acos%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bhypotenuse%7D%0A%5Cquad%20%0A%25%20tangent%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
![\bf sin(\theta )=\cfrac{2}{7}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{let's find the adjacent side} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{7^2-2^2}=a\implies \pm\sqrt{45}=a\implies \pm 3\sqrt{5}=a](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B2%7D%7B7%7D%5Ccfrac%7B%5Cleftarrow%20opposite%7D%7B%5Cleftarrow%20hypotenuse%7D%5Cqquad%20%5Ctextit%7Blet%27s%20find%20the%20adjacent%20side%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ac%3Dhypotenuse%5C%5C%0Aa%3Dadjacent%5C%5C%0Ab%3Dopposite%5C%5C%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%5Csqrt%7B7%5E2-2%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B45%7D%3Da%5Cimplies%20%5Cpm%203%5Csqrt%7B5%7D%3Da)
but.... which is it? the + or the -? well, we know that tan(θ) > 0, is another way to say that the tangent of the angle is positive, now, for the tangent to be positive, since it's opposite/adjacent both opposite and adjacent have to be the same exact sign, now, we know the opposite is +2, so that means the adjacent has to be the same sign, thus is the positive version 3√(5)
thus
ANSWER
![x=16;x=26](https://tex.z-dn.net/?f=x%3D16%3Bx%3D26)
EXPLANATION
We want to solve the equation:
![(x-21)^2=25](https://tex.z-dn.net/?f=%28x-21%29%5E2%3D25)
To do this, find the square root of both sides of the equation:
![\sqrt[]{(x-21)^2}=\sqrt[]{25}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%28x-21%29%5E2%7D%3D%5Csqrt%5B%5D%7B25%7D)
This will yield two results:
![\begin{gathered} \Rightarrow x-21=5 \\ \Rightarrow x-21=-5 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5CRightarrow%20x-21%3D5%20%5C%5C%20%5CRightarrow%20x-21%3D-5%20%5Cend%7Bgathered%7D)
Solving the two equations, we have:
![\begin{gathered} \Rightarrow x=5+21 \\ x=26 \\ \Rightarrow x=-5+21 \\ x=16 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5CRightarrow%20x%3D5%2B21%20%5C%5C%20x%3D26%20%5C%5C%20%5CRightarrow%20x%3D-5%2B21%20%5C%5C%20x%3D16%20%5Cend%7Bgathered%7D)
The solutions are:
Answer:
It would be (1,6), (-2,3)
x=1, y=6
x=-2, y=3
Step-by-step explanation:
Solve for the first variable in one of the equations, then substitute the result into the other equation.