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andrew-mc [135]
3 years ago
8

A car travels 60 kilometers in one hour before a piston breaks, then travels at 30 kilometers per hour for the remaining 60 kilo

meters to its destination. What is its average speed in kilometers per hour for the entire trip?
Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
3 0

Answer:

Total Distance : 1*60 +60=120

Total time taken = 1+ 60/30= 1+2=3

Hence average speed for the trip = 120/3= 40 kmph

Hence Answer is 40

Step-by-step explanation:

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Given two points (x₁,y₁) and (x₂,y₂), the slope of the line passes through these points will be:
m=(y₂-y₁)/(x₂-x₁)

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m=(-3+4)/(1+5)=1/6

Answer: the slope would be 1/6
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What is the value of the 4 in the number 27.43? write your answer as a fraction
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Answer:

4 tenths or 4/10

Step-by-step explanation:

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6 0
4 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
Which are steps in the process of completing the square used to solve the equation 3 – 4x = 5x2 – 14x? Check all that apply.
Bond [772]

Answer:

1. 3 = 5x2 – 10x

2. 8 = 5(x2 – 2x + 1)

3. StartFraction 8 Over 5 EndFraction = (x – 1)2

Step-by-step explanation:

3-4x=5x^2-14x

3=5x^2-14x+4x

3=5x^2-10x

5x^2-10x-3=0

1. 3 = 5x2 – 10x

2. 8 = 5(x2 – 2x + 1)

8=5x^2-10x+5

8-5=5x^2-10x

3=5x^2-10x

3. StartFraction 8 Over 5 EndFraction = (x – 1)2

8/5=x^2-2x+1

Cross product

8=5(x^2-2x+1)

8=5x^2-10x+5

8-5=5x^2-10x

3=5x^2-10x

7 0
3 years ago
Read 2 more answers
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