The answer for:
20)
6x-7+6x³-4x-10x³-4-8x - 2x²= −43−22−6−11
19)
8u+6+7u+(-10) + (-5u) + 8= 10+4
Answer: 5 km walking and 30 km by bus
Step-by-step explanation:
Yochanan walked from home to the bus stop at an average speed of 5 km / h. He immediately got on his school bus and traveled at an average speed of 60 km / h until he got to school. The total distance from his home to school is 35 km, and the entire trip took 1.5 hours. How many km did Yochanan cover by walking and how many did he cover by travelling on the bus?
walking - 5km/h bus - 60km/h
distance walking - d₁ distance bus - d₂
time walking - t₁ time bus - t₂
d₁ + d₂ = 35
t₁ + t₂ = 1.5
v = d/t
vwalking = d₁/t₁
5 = d₁/t₁ ⇒ d₁ = 5t₁
vbus = d₂/t₂
60 = d₂/t₂ ⇒ d₂ = 60t₂
d₁ + d₂ = 35 ⇒ 5t₁ + 60t₂ = 35
_________________________
5t₁ + 60t₂ = 35
t₁ + t₂ = 1.5 (*-5)
5t₁ + 60t₂ = 35
-5t₁ -5t₂ = -7.5 (+)
__________________________
55t₂ = 27.5
t₂ = 27.5/55 = 0.5 h
t₁ + t₂ = 1.5 ⇒ t₁ = 1.5 - 0.5 = 1h
d₁ = 5t₁ ⇒ d₁ = 5.1 = 5 km
d₂ = 60t₂ ⇒ d₂ = 30.0.5 = 30 km
Simultaneous Equations:
(3x + 5y = 7)×(-3)
(4x + 3y = 2)×(5)
-9x - 15y = -21
20x+15y =10 (-15y and and+15y cancels out)
11x = -11
=
(11 and 11 cancels out)
x= -1
substitute the x we found into either original expressions
3(-1) + 5y = 7
-3 + 5y = 7
+3 -3+ 5y = 7+ 3
5y = 10
=
(5 and 5 cancels out)
y=2
Check it!
4(-1)+3(2) = 2
-4 + 6 = 2
2 = 2
3(-1) + 5(2) = 7
-3 + 10 = 7
7 = 7
x= -1y= 2
Answer:
J,L M,O R,H
Step-by-step explanation:
Answer:
5.75 meters
Step-by-step explanation:
Here we are told that , the trolley added to the stack , increases the length by 25 cms or 0.25 mts.
Hence every trolley stacked will add 0.25 mts to the trolley already stacked. However the first trolley in 20 trolleys will be occupying 1 mt only. Hence , rest of the 19 trolleys will be covering 0.25×19=4.75 mts.
Hence , total distance covered by 20 trolleys =
1+4.75=5.75 mts
