Answer:
An equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:
Step-by-step explanation:
We know that the slope-intercept form of the line equation is
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
where
Given the line
y = -13x+4
comparing with the slope-intercept form of the line equation
The slope of the line = m = -13
We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:
slope = m = -13
Thus, the slope of the the new perpendicular line = – 1/m = -1/-13 = 1/13
Using the point-slope form of the line equation
![y-y_1=m\left(x-x_1\right)](https://tex.z-dn.net/?f=y-y_1%3Dm%5Cleft%28x-x_1%5Cright%29)
where
- m is the slope of the line
substituting the values of the slope m = 1/13 and the point (4, 3)
![y-y_1=m\left(x-x_1\right)](https://tex.z-dn.net/?f=y-y_1%3Dm%5Cleft%28x-x_1%5Cright%29)
![y-3=\frac{1}{13}\left(x-4\right)](https://tex.z-dn.net/?f=y-3%3D%5Cfrac%7B1%7D%7B13%7D%5Cleft%28x-4%5Cright%29)
Add 3 to both sides
![y-3+3=\frac{1}{13}\left(x-4\right)+3](https://tex.z-dn.net/?f=y-3%2B3%3D%5Cfrac%7B1%7D%7B13%7D%5Cleft%28x-4%5Cright%29%2B3)
![y=\frac{1}{13}x-\frac{4}{13}+3](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B13%7Dx-%5Cfrac%7B4%7D%7B13%7D%2B3)
![y=\frac{1}{13}x+\frac{35}{13}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B13%7Dx%2B%5Cfrac%7B35%7D%7B13%7D)
Therefore, an equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:
So the first thing we need to do is set up an equation (using x to represent the number). Then, we'll solve it to get the answer.
Step 1: Translate the word problem into an equation
Word problem: The sum of nine and six-tenths and a number is thirteen and two-tenths.
Equation:
+ x = ![13 \frac{2}{10}](https://tex.z-dn.net/?f=13%20%5Cfrac%7B2%7D%7B10%7D)
Step 2: Isolate x by subtracting
from each side
+ x -
=
- ![9 \frac{6}{10}](https://tex.z-dn.net/?f=9%20%5Cfrac%7B6%7D%7B10%7D)
Step 3: The numbers on the left cancel out, leaving x alone
x = ![3 \frac{6}{10}](https://tex.z-dn.net/?f=3%20%5Cfrac%7B6%7D%7B10%7D)
Step 4: Convert the answer to decimals
x = 3.6
Answer:
1.0
Step-by-step explanation:
The answer is 1.0 It is not a tenth but 99 is closer to 100. witch is 1.0 Do you try looking it up. This is a site i used. It is great. It is called calculator soup.
Answer:
Part 1) ![x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-2x-2%3D%28x-1-%5Csqrt%7B3%7D%29%28x-1%2B%5Csqrt%7B3%7D%29)
Part 2) ![x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-6x%2B4%3D%28x-3-%5Csqrt%7B5%7D%29%28x-3%2B%5Csqrt%7B5%7D%29)
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
![ax^{2} +bx+c=0](https://tex.z-dn.net/?f=ax%5E%7B2%7D%20%2Bbx%2Bc%3D0)
is equal to
![x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%28%2B%2F-%29%5Csqrt%7Bb%5E%7B2%7D-4ac%7D%7D%20%7B2a%7D)
Part 1)
in this problem we have
![x^{2} -2x-2=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-2x-2%3D0)
so
![a=1\\b=-2\\c=-2](https://tex.z-dn.net/?f=a%3D1%5C%5Cb%3D-2%5C%5Cc%3D-2)
substitute in the formula
![x=\frac{-(-2)(+/-)\sqrt{-2^{2}-4(1)(-2)}} {2(1)}\\\\x=\frac{2(+/-)\sqrt{12}} {2}\\\\x=\frac{2(+/-)2\sqrt{3}} {2}\\\\x_1=\frac{2(+)2\sqrt{3}} {2}=1+\sqrt{3}\\\\x_2=\frac{2(-)2\sqrt{3}} {2}=1-\sqrt{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-2%29%28%2B%2F-%29%5Csqrt%7B-2%5E%7B2%7D-4%281%29%28-2%29%7D%7D%20%7B2%281%29%7D%5C%5C%5C%5Cx%3D%5Cfrac%7B2%28%2B%2F-%29%5Csqrt%7B12%7D%7D%20%7B2%7D%5C%5C%5C%5Cx%3D%5Cfrac%7B2%28%2B%2F-%292%5Csqrt%7B3%7D%7D%20%7B2%7D%5C%5C%5C%5Cx_1%3D%5Cfrac%7B2%28%2B%292%5Csqrt%7B3%7D%7D%20%7B2%7D%3D1%2B%5Csqrt%7B3%7D%5C%5C%5C%5Cx_2%3D%5Cfrac%7B2%28-%292%5Csqrt%7B3%7D%7D%20%7B2%7D%3D1-%5Csqrt%7B3%7D)
therefore
![x^{2} -2x-2=(x-(1+\sqrt{3}))(x-(1-\sqrt{3}))](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-2x-2%3D%28x-%281%2B%5Csqrt%7B3%7D%29%29%28x-%281-%5Csqrt%7B3%7D%29%29)
![x^{2} -2x-2=(x-1-\sqrt{3})(x-1+\sqrt{3})](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-2x-2%3D%28x-1-%5Csqrt%7B3%7D%29%28x-1%2B%5Csqrt%7B3%7D%29)
Part 2)
in this problem we have
![x^{2} -6x+4=0](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-6x%2B4%3D0)
so
![a=1\\b=-6\\c=4](https://tex.z-dn.net/?f=a%3D1%5C%5Cb%3D-6%5C%5Cc%3D4)
substitute in the formula
![x=\frac{-(-6)(+/-)\sqrt{-6^{2}-4(1)(4)}} {2(1)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-6%29%28%2B%2F-%29%5Csqrt%7B-6%5E%7B2%7D-4%281%29%284%29%7D%7D%20%7B2%281%29%7D)
![x=\frac{6(+/-)\sqrt{20}} {2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B6%28%2B%2F-%29%5Csqrt%7B20%7D%7D%20%7B2%7D)
![x=\frac{6(+/-)2\sqrt{5}} {2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B6%28%2B%2F-%292%5Csqrt%7B5%7D%7D%20%7B2%7D)
![x_1=\frac{6(+)2\sqrt{5}}{2}=3+\sqrt{5}](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B6%28%2B%292%5Csqrt%7B5%7D%7D%7B2%7D%3D3%2B%5Csqrt%7B5%7D)
![x_2=\frac{6(-)2\sqrt{5}}{2}=3-\sqrt{5}](https://tex.z-dn.net/?f=x_2%3D%5Cfrac%7B6%28-%292%5Csqrt%7B5%7D%7D%7B2%7D%3D3-%5Csqrt%7B5%7D)
therefore
![x^{2} -6x+4=(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-6x%2B4%3D%28x-%283%2B%5Csqrt%7B5%7D%29%29%28x-%283-%5Csqrt%7B5%7D%29%29)
![x^{2} -6x+4=(x-3-\sqrt{5})(x-3+\sqrt{5})](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-6x%2B4%3D%28x-3-%5Csqrt%7B5%7D%29%28x-3%2B%5Csqrt%7B5%7D%29)