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s2008m [1.1K]
3 years ago
10

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 51 km/h. After five hou

rs, the velocity of the car is 59 km/h.
Part A: Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used.

Part B: How can you graph the equation obtained in Part A for the first six hours?
Mathematics
1 answer:
Alex777 [14]3 years ago
5 0
The question is asking to write and equation by using the two variable in the standard form that ca be used to describe the velocity of thecar at different times and base on my calculation, I would say that the answer would be v=4t+39. I hope you are satisfied with my answer and feel free to ask for more 
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Value of x<br><br>30x=30-30​
Vladimir79 [104]

Answer:

<h2>0</h2>

Step-by-step explanation:

30x = 30 - 30

=  > 30x = 0

=  > x =  \frac{0}{30}

=  > x = 0

5 0
2 years ago
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A line has a slope of 9. One of the points on the line is (-2, p). Another point on the line is (2, 2p+1). Determine the value o
igomit [66]

As we travel from (-2, p) to (2, 2p+1), x increases by 4 and y increases by p+1.  Note that -2 + 4 = 2 (which is correct), and that p + (p+1) = 2p+1 (which is also correct).                                                        

                                                                           p+1

Thus, the slope of this line is m = rise / run = -------- = 9

                                                                               4


Then p + 1 = 36, and p = 35.

8 0
3 years ago
Recall that m(t) = (1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500g sample of phosphorus-32 decays
katrin2010 [14]

The question is incomplete, here is the complete question:

Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

<u>Answer:</u> The half life of the sample of phosphorus-32 is 14.28days^{-1}

<u>Step-by-step explanation:</u>

The equation used to calculate the half life of the sample is given as:

m(t)=m_o(1/2)^{t/h}

where,

m(t) =  amount of sample after time 't' = 356 g

m_o = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}

Hence, the half life of the sample of phosphorus-32 is 14.28days^{-1}

7 0
3 years ago
an aquarium holds 33.75 gallons of water. it has a length of 2 feet and a height of 1.5 feet what is the width of the aquarium
kolezko [41]
33.75 = 2*1.5*x
33.75 = 3x
x = 11.25 feet
6 0
3 years ago
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Edward is driving a car on the highway. He drives for 28.5 miles in 3/8 of an hour. At this rate, how much distance can he cover
kari74 [83]

Answer:

76 Miles.

Step-by-step explanation:

28.5/3=9.5

9.5(8)=76

5 0
3 years ago
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