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dybincka [34]
3 years ago
11

The length of a rectangle is increasing at a rate of 8cm/s and its width is increasing at a rate of 3 cm/s. When the length is 2

0 cm and the width is 10 cm, how fast is the area of the rectangle increasing?
Mathematics
1 answer:
Aloiza [94]3 years ago
8 0

Answer:

Step-by-step explanation:

let length=l

width=w

Area A=lw

\frac{dA}{dt}=l \frac{dw}{dt}+w \frac{di}{dt}\\\frac{dl}{dt}=8 cm/s\\\frac{dw}{dt}= 3 cm/s\\\frac{dA}{dt}=20 \times 3+10 \times 8=60+80=140 cm ^{2} /s

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7 0
3 years ago
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
kari74 [83]

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

6 0
2 years ago
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