Dividing by a fraction is equivalent to multiply by its reciprocal, then:
Now, we need to express the quadratic polynomials using their roots, as follows:
where y1 and y2 are the roots.
Applying the quadratic formula to the first polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B7%5Cpm%5Csqrt%5B%5D%7B%28-7%29%5E2-4%5Ccdot3%5Ccdot%28-6%29%7D%7D%7B2%5Ccdot3%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B7%5Cpm%5Csqrt%5B%5D%7B121%7D%7D%7B6%7D%20%5C%5C%20y_1%3D%5Cfrac%7B7%2B11%7D%7B6%7D%3D3%20%5C%5C%20y_2%3D%5Cfrac%7B7-11%7D%7B6%7D%3D-%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the second polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B1%5E2-4%5Ccdot2%5Ccdot%28-3%29%7D%7D%7B2%5Ccdot2%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B25%7D%7D%7B4%7D%20%5C%5C%20y_1%3D%5Cfrac%7B-1%2B5%7D%7B4%7D%3D1%20%5C%5C%20y_2%3D%5Cfrac%7B-1-5%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the third polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B3%5Cpm%5Csqrt%5B%5D%7B%28-3%29%5E2-4%5Ccdot2%5Ccdot%28-9%29%7D%7D%7B2%5Ccdot2%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B3%5Cpm%5Csqrt%5B%5D%7B81%7D%7D%7B4%7D%20%5C%5C%20y_1%3D%5Cfrac%7B3%2B9%7D%7B4%7D%3D3%20%5C%5C%20y_2%3D%5Cfrac%7B3-9%7D%7B4%7D%3D-%5Cfrac%7B3%7D%7B2%7D%20%5Cend%7Bgathered%7D)
Applying the quadratic formula to the fourth polynomial:
![\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y_%7B1%2C2%7D%3D%5Cfrac%7B-b%5Cpm%5Csqrt%5B%5D%7Bb%5E2-4ac%7D%7D%7B2a%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B1%5E2-4%5Ccdot1%5Ccdot%28-2%29%7D%7D%7B2%5Ccdot1%7D%20%5C%5C%20y_%7B1%2C2%7D%3D%5Cfrac%7B-1%5Cpm%5Csqrt%5B%5D%7B9%7D%7D%7B2%7D%20%5C%5C%20y_1%3D%5Cfrac%7B-1%2B3%7D%7B2%7D%3D1%20%5C%5C%20y_2%3D%5Cfrac%7B-1-3%7D%7B2%7D%3D-2%20%5Cend%7Bgathered%7D)
Substituting into the rational expression and simplifying:
This would be 13+2x because if x is the number, then 2 times it would be 2x and the sum means addition so it would be 13+2x
7*-4= -28 +4-12/2
-24-6
=-30
Simple Pythagorean theorem, And if you double check Tamara's work she is correct. Therefor B is the correct answer.
The equation of the line that is perpendicular to y = -5/2 x + 1 and passing through (10, 3) will be y = 2/5 x + 1.
<h3>What is the linear system?</h3>
A linear system is one in which the parameter in the equation has a degree of one. It might have one, two, or even more variables.
The equation of the perpendicular line will be given as
y = mx + c
The given equation is y = -5/2 x + 1.
The slope of the line is -5/2.
Then the product of the slope of the perpendicular lines is a negative one.
-5/2 × m = - 1
m = 2/5
Then the equation will be
y = 2/5 x + c
Then the equation is passing through (10, 3). Then we have
3 = 2/5 × 10 + c
3 = 2 + c
c = 1
Then the equation is y = 2/5 x + 1.
More about the linear system link is given below.
brainly.com/question/20379472
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