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Aliun [14]
3 years ago
6

Which equation has the components of 0 = x2 – 9x – 20 inserted into the quadratic formula correctly?

Mathematics
3 answers:
snow_lady [41]3 years ago
8 0

Answer:

This is the right answer

x = StartFraction 9 plus or minus StartRoot (negative 9) squared minus 4(1)(negative 20) EndRoot Over 2(1) EndFraction

Aneli [31]3 years ago
6 0

Answer:

The required equation is x=\frac{-(-9)\pm \sqrt{(-9)^2-4(1)(-20)}}{2(1)}.

Step-by-step explanation:

If a quadratic equation is ax^2+bx+c=0, then the quadratic formula is

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

The given quadratic equation is

x^2-9x-20=0

Here, a=1, b=-9 and c=-20.

Substitute a=1, b=-9 and c=-20 in the above quadratic formula.

x=\frac{-(-9)\pm \sqrt{(-9)^2-4(1)(-20)}}{2(1)}

Therefore the required equation is x=\frac{-(-9)\pm \sqrt{(-9)^2-4(1)(-20)}}{2(1)}.

XxMandy_WolfxX2 years ago
0 0

It's C on edge 2021

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4x + 8y=20<br> 3x+6y=15<br> solve this system of equations
laila [671]
4x=20-8y

x=5-2y now use this value of x in the second equation...

3(5-2y)+6y=15

15-6y+6y=15

15=15  This is true for any y or x value.

So there are infinitely many solutions as the two equations describe the same line.
5 0
3 years ago
Read 2 more answers
What is the slope of (4,1) and (6,-1)
schepotkina [342]

Answer:

slope = -1

Step-by-step explanation:

you <u>subtract the y-values</u> (-1-1=-2) then <u>subtract the x-values</u> (6-4=2) then divide the sum of the y-values by the sum of the x-values

\frac{-2}{2} which simplifies to a slope of -1

6 0
2 years ago
What is the area of a regular octagon with a side length of 4.6 meters and a length from the center to a vertex of 6 meters? HEL
Shtirlitz [24]

<span>What is the area of a regular octagon with a side length of 4.6 meters and a length from the center to a vertex of 6 meters? 

Good Morning
 
The Correct Answer is c) 101.2

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</span>
6 0
3 years ago
If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle
stira [4]

Answer:

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<u>Step-by-step explanation:</u>

Given points end  Points are p(-3,-2) and q( 1,6)

<em>The distance of two points formula</em>

P Q = \sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2}   }

P Q = \sqrt{1 - (-3)^{2} + ((6 -(-2))^{2}   }

P Q = \sqrt{16+64} = \sqrt{80}

<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

<em>                        </em>(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )<em></em>

<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

6 0
4 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
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