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Usimov [2.4K]
3 years ago
15

30 POINTS!!! Hello!! Marie here! Im giving BRAINLIEST and THANKS and FIVE STARS to the person who answers this question and expl

ains it!
How much miles is race track if the racers did 500 miles in 200 laps?
Mathematics
1 answer:
9966 [12]3 years ago
5 0
<h2>.....Question:-)</h2>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>How </em><em>much </em><em>miles </em><em>is </em><em>race </em><em>track </em><em>if </em><em>the </em><em>racers </em><em>did </em><em>5</em><em>0</em><em>0</em><em>m</em><em>i</em><em>l</em><em>e</em><em>s</em><em> </em><em>in </em><em>2</em><em>0</em><em>0</em><em> </em><em>laps?</em><em>?</em>

<h2><em>Answer</em><em> (✷‿✷)</em></h2>

<em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>2</em><em>.</em><em>5</em><em> </em><em>miles </em><em>(</em><em>4</em><em>k</em><em>m</em><em>)</em><em> </em><em>is </em><em>race </em><em>track </em><em>if </em><em>the </em><em>racers </em><em>did </em><em>5</em><em>0</em><em>0</em><em> </em><em>miles </em><em>in </em><em>2</em><em>0</em><em>0</em><em> </em><em>laps.</em><em>.</em><em>.</em><em>.</em>

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What is the distance between (5,8) and (2,4) by using distance formula
Dimas [21]

Answer:

Distance = 5

Step-by-step explanation:

D= sqrt (x1-x2)^2 + (y1-y2)^2

(5-2)^2 + (8-4)^2=3^2 + 4 ^2 = 9 +16 = 25

Sqrt of 25 = 5

Distance = 5

Hope this helps

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3 years ago
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6 0
3 years ago
Solve the inequality a/5 less than 1.
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Answer:

hummmm

Step-by-step explanation:

Most of the time, an inequality has more than one or even infinity solutions. For example the inequality: x>3 . The solutions of this inequality are "all numbers strictly greater than 3". ... The inequality has an infinite amount of solutions

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3 years ago
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X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square
swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = -\frac{4}{5}x+\frac{58}{5}

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

   [x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

   (x + 2)² + (y - 5)²- 29 + 20 = 30

   (x + 2)² + (y - 5)²- 9 = 30

   (x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

    Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

    Slope of the line joining this point to the center (-2, 5),

    m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

          = \frac{10-5}{2+2}

          = \frac{5}{4}

    Let the slope of the tangent which is perpendicular to this line is 'm_{2}'

    Then by the property of perpendicular lines,

          m_{1}\times m_{2}=-1

          \frac{5}{4}\times m_{2}=-1

                 m_{2}=-\frac{4}{5}

   Now the equation of the line passing though (2, 10) having slope m_{2}=-\frac{4}{5}

           y - y' = m_{2}(x-x')

           y - 10 = -\frac{4}{5}(x-2)

           y - 10 = -\frac{4}{5}x+\frac{8}{5}

                  y = -\frac{4}{5}x+\frac{8}{5}+10

                  y = -\frac{4}{5}x+\frac{58}{5}

Therefore, equation of the line will be, y = -\frac{4}{5}x+\frac{58}{5}

7 0
3 years ago
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