Answer:
![\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}](https://tex.z-dn.net/?f=%5Cmathrm%7BThe%5C%3Asolutions%5C%3Ato%5C%3Athe%5C%3Aquadratic%5C%3Aequation%5C%3Aare%3A%7D)
![x=3,\:x=-\frac{1}{2}](https://tex.z-dn.net/?f=x%3D3%2C%5C%3Ax%3D-%5Cfrac%7B1%7D%7B2%7D)
Step-by-step explanation:
considering the equation
![2x^2\:-\:4x\:-\:3\:=\:x](https://tex.z-dn.net/?f=2x%5E2%5C%3A-%5C%3A4x%5C%3A-%5C%3A3%5C%3A%3D%5C%3Ax)
solving
![2x^2\:-\:4x\:-\:3\:=\:x](https://tex.z-dn.net/?f=2x%5E2%5C%3A-%5C%3A4x%5C%3A-%5C%3A3%5C%3A%3D%5C%3Ax)
![2x^2-4x-3-x=x-x](https://tex.z-dn.net/?f=2x%5E2-4x-3-x%3Dx-x)
![2x^2-5x-3=0](https://tex.z-dn.net/?f=2x%5E2-5x-3%3D0)
![\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3Aa%5C%3Aquadratic%5C%3Aequation%5C%3Aof%5C%3Athe%5C%3Aform%5C%3A%7Dax%5E2%2Bbx%2Bc%3D0%5Cmathrm%7B%5C%3Athe%5C%3Asolutions%5C%3Aare%5C%3A%7D)
![x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
![\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7D%5Cquad%20a%3D2%2C%5C%3Ab%3D-5%2C%5C%3Ac%3D-3%3A%5Cquad%20x_%7B1%2C%5C%3A2%7D%3D%5Cfrac%7B-%5Cleft%28-5%5Cright%29%5Cpm%20%5Csqrt%7B%5Cleft%28-5%5Cright%29%5E2-4%5Ccdot%20%5C%3A2%5Cleft%28-3%5Cright%29%7D%7D%7B2%5Ccdot%20%5C%3A2%7D)
solving
![x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%5Cleft%28-5%5Cright%29%2B%5Csqrt%7B%5Cleft%28-5%5Cright%29%5E2-4%5Ccdot%20%5C%3A2%5Cleft%28-3%5Cright%29%7D%7D%7B2%5Ccdot%20%5C%3A2%7D)
![x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%2B%5Csqrt%7B%5Cleft%28-5%5Cright%29%5E2%2B4%5Ccdot%20%5C%3A2%5Ccdot%20%5C%3A3%7D%7D%7B2%5Ccdot%20%5C%3A2%7D)
![x=\frac{5+\sqrt{49}}{2\cdot \:2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%2B%5Csqrt%7B49%7D%7D%7B2%5Ccdot%20%5C%3A2%7D)
![x=\frac{5+7}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5%2B7%7D%7B4%7D)
![x=3](https://tex.z-dn.net/?f=x%3D3)
also solving
![x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%5Cleft%28-5%5Cright%29-%5Csqrt%7B%5Cleft%28-5%5Cright%29%5E2-4%5Ccdot%20%5C%3A2%5Cleft%28-3%5Cright%29%7D%7D%7B2%5Ccdot%20%5C%3A2%7D)
![x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5-%5Csqrt%7B%5Cleft%28-5%5Cright%29%5E2%2B4%5Ccdot%20%5C%3A2%5Ccdot%20%5C%3A3%7D%7D%7B2%5Ccdot%20%5C%3A2%7D)
![x=\frac{5-\sqrt{49}}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B5-%5Csqrt%7B49%7D%7D%7B4%7D)
![x=-\frac{2}{4}](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7B2%7D%7B4%7D)
![x=-\frac{1}{2}](https://tex.z-dn.net/?f=x%3D-%5Cfrac%7B1%7D%7B2%7D)
Therefore,
![\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}](https://tex.z-dn.net/?f=%5Cmathrm%7BThe%5C%3Asolutions%5C%3Ato%5C%3Athe%5C%3Aquadratic%5C%3Aequation%5C%3Aare%3A%7D)
![x=3,\:x=-\frac{1}{2}](https://tex.z-dn.net/?f=x%3D3%2C%5C%3Ax%3D-%5Cfrac%7B1%7D%7B2%7D)
Answer:
x = 6
y = 2
Step-by-step explanation:
Answer:
![y=-\frac{1}{3}x+7](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B3%7Dx%2B7)
Step-by-step explanation:
we know that
If two lines are parallel then their slopes are equal
The equation of the given line is
![y=-\frac{1}{3}x+4](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B3%7Dx%2B4)
the slope is ![m=-\frac{1}{3}](https://tex.z-dn.net/?f=m%3D-%5Cfrac%7B1%7D%7B3%7D)
so
the slope of the parallel line to the given line is also ![m=-\frac{1}{3}](https://tex.z-dn.net/?f=m%3D-%5Cfrac%7B1%7D%7B3%7D)
Find the equation of the line that is parallel to the given line and passes through the point (6, 5)
we have
![m=-\frac{1}{3}](https://tex.z-dn.net/?f=m%3D-%5Cfrac%7B1%7D%7B3%7D)
![point\ (6,5)](https://tex.z-dn.net/?f=point%5C%20%286%2C5%29)
The equation of the line in point slope form is
![y-y1=m(x-x1)](https://tex.z-dn.net/?f=y-y1%3Dm%28x-x1%29)
substitute
![y-5=-\frac{1}{3}(x-6)](https://tex.z-dn.net/?f=y-5%3D-%5Cfrac%7B1%7D%7B3%7D%28x-6%29)
Convert to slope intercept form
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
isolate the variable y
![y-5=-\frac{1}{3}x+2](https://tex.z-dn.net/?f=y-5%3D-%5Cfrac%7B1%7D%7B3%7Dx%2B2)
![y=-\frac{1}{3}x+2+5](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B3%7Dx%2B2%2B5)
![y=-\frac{1}{3}x+7](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B1%7D%7B3%7Dx%2B7)
Answer:
5 sqrt(10) = c
Step-by-step explanation:
Since this is a right triangle, we can use the Pythagorean theorem
a^2 + b^2 = c^2
9^2 + 13^2 = c^2
81 + 169 = c^2
250 = c^2
Take the square root of each side
sqrt(250) = c
sqrt(25) sqrt(10) = c
5 sqrt(10) = c
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