Question #210711

Exercise 102


Suppose a ball is dropped from a height of 200m. If time after the ball was released is in seconds, and the height of the ball above the ground at time is given by

 , find the average velocity from to 


Show that the slope of the line joining the points A(1,1) and B is 


A line passes through the points (0, 5) and (9, −1). Find the equation of the line which is perpendicular to the line and passes through its midpoint.


Find the equation of the line that is parallel to the line y = −2x + 6 and passing through the point A(1, 10).


A square has vertices O(0, 0), A(a, 0), B(a, a) and C(0, a).

i Find the midpoint of the diagonals OB and CA.

ii Find the length of a diagonal of the square and the radius of the circle in which OABC is inscribed.

Iii Find the equation of the circle inscribing the square




1
Expert's answer
2022-02-22T08:47:46-0500

1. h(t)=20010t2h(t)=200-10t^2


h(2)=20010(2)2=160h(2)=200-10(2)^2=160

h(3)=20010(3)2=110h(3)=200-10(3)^2=110

The average velocity from 2s to 3s is


vave=110m160m3s2s=50m/sv_{ave}=\dfrac{110m-160m}{3s-2s}=-50m/s

2.


A(1,1),B(2,3+h)A(1, 1), B(2, 3+h)

slope=yByAxBxA=3+h121=2+hslope=\dfrac{y_B-y_A}{x_B-x_A}=\dfrac{3+h-1}{2-1}=2+h

3. Midpoint (9/2,2)(9/2, 2)


slope1=1590=23slope_1=\dfrac{-1-5}{9-0}=-\dfrac{2}{3}

If line is perpendicular then


slope2=1/(23)=32slope_2=-1/(-\dfrac{2}{3})=\dfrac{3}{2}

y=32x+by=\dfrac{3}{2}x+b

2=32(92)+b=>b=1942=\dfrac{3}{2}(\dfrac{9}{2})+b=>b=-\dfrac{19}{4}

The equation of the line is


y=32x194y=\dfrac{3}{2}x-\dfrac{19}{4}

4.


y=2x+by=-2x+b

10=2(1)+b=>b=1210=-2(1)+b=>b=12

The equation of the line is


y=2x+10y=-2x+10

5.

i.


M(a/2,a/2)M(a/2, a/2)

ii.


AC=BD=a2,>0AC=BD=a\sqrt{2}, >0

r=a/2,a>0r=a/2, a>0

iii.


R=OB/2=2aa,a>0R=OB/2=\dfrac{\sqrt{2}a}{a}, a>0


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