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steposvetlana [31]
3 years ago
6

Find the point on the line y = 3x + 2 that is closest to the origin.

Mathematics
1 answer:
anzhelika [568]3 years ago
3 0
1.
the slope of the line y=3x+2 is the coefficient of x, that is 3.

2.
the slope of any line perpendicular to the line y=3x+2, is m, such that:

3*m=-1

so m= \frac{-1}{3}

3. 
so any line with equation y=-1/3x+k is perpendicular to y=3x+2.

moreover, the line y=-1/3x (so k=0) is the line passing through the origin
(0, 0) and perpendicular to y=3x+2


4. This means that the closest point of y=3x+2 to the origin is the intersection point of lines y=3x+2 and y=-1/3x

5.
to find this point:

3x+2=(-1/3)x
x(3+1/3)=-2
x(9/3+1/3)=-2

x*10/3=-2

x= \frac{-2*3}{10} = \frac{-6}{10}=-0.6


thus y is :  (-1/3)(-6/10)=0.2



Answer: (-0.6, 0.2)

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taurus [48]

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r^3=\dfrac{\frac{63}{8}}{\frac{7}{3}}\\\\r^3=\dfrac{63}{8}\cdot\dfrac{3}{7}\\\\r^3=\dfrac{9\cdot3}{8\cdot1}\\\\r^3=\dfrac{27}{8}\to r=\sqrt[3]{\dfrac{27}{8}}\\\\r=\dfrac{3}{2}\\\\h=2\dfrac{1}{3}\cdot\dfrac{3}{2}=\dfrac{7}{3}\cdot\dfrac{3}{2}=\dfrac{7}{2}\\\\k=\dfrac{7}{2}\cdot\dfrac{3}{2}=\dfrac{21}{4}\\\\Answer:\ \boxed{h=\dfrac{7}{2}=3\dfrac{1}{2}\ and\ k=\dfrac{21}{4}=5\dfrac{1}{4}}

8 0
2 years ago
4. What kind of triangle is made by connecting the points A(0, –6), B(3, –6), and C(3, –2)?
Firdavs [7]

Answer:

Part 4) Right triangle

Part 5) Kite

Step-by-step explanation:

Part 4) What kind of triangle is made by connecting the points A(0, –6), B(3, –6), and C(3, –2)?

Using a graphing tool    

see the attached figure N 1

The triangle of the figure is not equilateral------> The triangle does not have three equal sides

The triangle of the figure is a right triangle------>The triangle  has an angle of 90\°

The triangle of the figure is not isosceles------> The triangle does not have two equal sides

The triangle of the figure is not a right and isosceles

Part 5) What type of quadrilateral is formed by connecting the points (0, 9), (3, 6), (0, 1), and (-3, 6)?                

Using a graphing tool    

see the attached figure N 2        

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2 years ago
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frosja888 [35]

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Step-by-step explanation:

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2 years ago
Solve this application problem using a system of equations: The Bright College footballteam scored 80 times last season, some on
Goryan [66]

Answer:

\large \boxed{\sf \ \ 69 \ \textbf{touchdowns } 11 \ \textbf{field goals } \ }

Step-by-step explanation:

Hello,

Touchdown is 7 points, field goals is 3 points

Let's note a the number of touchdowns and b the number of field goals, we can write

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<=> 4a = 276

<=> a=\dfrac{276}{4}=69

And then from (1) b = 80 - 69 = 11

Hope this helps

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