Answer:
Given speed of trolley = 125 meters per minute
Now we know that
Distance= Speed x time
Thus for the first case since the time of travel is less than 450 seconds thus the distance traveled is less than
Distance < 2.083 x 450 =937.5 meters
hence depending on the given information we cannot come to any conclusion weather the distance travelled is less than 800 m or greater than 800 m.
For the second case
since the time of travel is greater than 400 seconds
Thus the distance traveled is
which is greater than 800 meters.
If your just figuring out how much the car depreciated just by driving off the lot new...1st year.
$30,000×.30=$9,000
$30,000- $9,000=$21,000.
just take the brand new value of the car multiplied by the .30% of the 1st depreciation...take that answer ($9,000) and subtract it from your orginal value of $31,000.
which gives you $21,000
Answer:
x=151/64 y=9/8 z=-51/32
Step-by-step explanation:
2x+2y+5z- (2x-y+z)=-1-2
3y+4z=-3
2x+2y+5z- (2x+4y-3z)=-1-14
-2y+8z=-15
We have two equations 3y+4z=-3 and -2y+8z=-15 (We get it due to elimination method)
3y+4z=-3 (2)
-2y+8z=-15 (-3)
6y+8z=-6
6y-24z=45
(6y+8z)- (6y-24z)= -6-45
32z=-51
z=-51/32
3y-204/32=-3
3y= -96/32+204/32
3y=108/32
y=36/32=9/8
2x+2*9/8+5*(-51/32)=-1
2x+9/4-255/32=-1
2x+72/32-255/32=-1
2x-183/32=-32/32
2x=151/32
x=151/64
Answer:
the question is incomplete, below is the complete question
"ress the following complex numbers in rectangular form. Show how you get the answer and use a calculator to verify your answer. E.g. 2 pts for 2∠30°=2(cos30°+jsin30°)=1.73+j, 1 pt for 2∠30°=1.73+j. Same grading criteria as 1.4. (a) Z1=5eⁱ³⁰ (b) z2=−3 ∠(−45°) (c) z3=2∠(−90°)
answer
a.Z1=4.33+j2.5
b. Z2=-2.12+j2.12
c.Z3=-2j
Step-by-step explanation
note that
Z=reⁱⁿ=r(cosπ+jsinπ)
hence from Z1=5eⁱ³⁰ wen have
Z1=5(cos30+jsin30)
Z1=5(0.8660+j0.5)
Z1=4.33+j2.5
b.also from z=r∠(π)=r(cosπ+jsinπ)
Hence,
z2=−3 ∠(−45°)=-3(cos(-45)+jsin(-45))
Z2=-3(0.7071-j0.7071)
Z2=-2.12+j2.12
c. z3=2∠(−90°)=2(cos(-90)+jsin(-90))
Z3=2(0-j)
Z3=-2j
