Answer:
2000? ... assume $2000
Resale Value = $2000*(1 - 0.25)t/yr
where: t = number of year after purchase
at t = 3 yr
Resale Value = $2000*(1 - 0.25)3 = $843.75
checking: at t = 0 $2000 (purchase price)
at t = 1 yr $2000 - 0.25*$2000 = $2000 - $500 = $1500
at t = 2 yr $1500 - 0.25*$1500 = $1500 - $375 = $1125
at t = 3 yr $1125 - 0.25*$1125 = $1125 - $281.25 = $843.75
Absolute value is how far away a number is from zero on the number line so -32 and 32 have the same absolute value and for number 2 i agree necause 30 is closer to zero than 32
Answer:
Step-by-step explanation:
Given: The radius of circle O is r, and the radius of circle X is r'.
To prove: Circle O is similar to circle X.
Proof: Move the center of the smaller circle onto the center of the largest circle. Translate the circle X by the vector XA onto circle O. The circles now have the same center.
A dilation is needed to increase the size of circle X to coincide with the circle O. A value which when multiplied by r' will create r.
The scale factor x to increase X:
⇒
A translation followed by a dilation with scale factor will map one circle to the other, thus proving the given both circles similar.
Therefore, circle O is similar to circle X.
Step-by-step explanation:
To find the answer, we can first find the distance each of them have, then divide it by their speed to find the time needed.
Distance Steve need to travel: 300 feet
Distance Paula need to travel:
300+175(behind steve) = 475 feet
Time needed respectively:
Steve: 300 ÷ 9 = 33.33.. seconds
Paula: 475 ÷ 15 = 31.66... seconds
As we can see from the result, Paula would take less time to reach the goal (31.66<33.33),
therefore, Paula win the bike race by:
33.33-31.67
=1.66...
≈1.67 feet
Thus Paula won by 1.67 feet.
Hope it helps!
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)
We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>
The vector product pq x pr gives a vector perpendicular to both pq and pr. This vector is the normal vector of a plane passing through all three points
pq x pr
=
i j k
-4 -2 -4
-3 5 1
=<-2+20,12+4,-20-6>
=<18,16,-26>
Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>
The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).
The equation of the required plane is therefore
Π : 9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π : 9x+8y-13z=24
Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.
