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3 years ago

The table shows the distance traveled by an object moving at a constant speed. How far will the object go in 1 minute?

Mathematics

2 answers:

kiruha [24]3 years ago

8 0

Answer: 510 meters

Step-by-step explanation:

Given: The table shows the distance traveled by an object moving at a constant speed.

We know that the $\text{speed}=\frac{\text{distance}}{\text{time}}$

From table in 10 seconds the distance traveled = 85 meters

$\\\\\Rightarrow\text{speed}=\frac{85}{10}=8.5$ meters per second

If Time = 1 minutes

Also, 1 minute=60 seconds

Since, speed is constant , then the distance traveled in 1 min is given by :_

$\text{distance}=\text{speed}\times\text{time}\\\Rightarrow\text{distance}=8.5\times60=510\text{ meters}$

Ronch [10]3 years ago

4 0

If the object moves 170 meters in 20 seconds. There are 60 seconds in a minute. You have to multiply the 170 by 3. One minute it moves 510 meters

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A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the

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(a) Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T - 75)$ where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get $\dfrac{dT}{T - 75} = k dt$ . Integrate both sides to get $\ln|T - 75| = kt + C$ .

Since $T(0) = 185$ , we solve for C:
$|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110$
So we get $\ln|T - 75| = kt + \ln 110$ . Use T(30) = 150 to solve for k:
$\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)$

So

$\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t} \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}$

But choose Positive because T > 75. Temp of turkey can't go under.

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)} = 136.929 \approx 137{}^{\circ}F$

(b)

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30} \\
100 = 75 + 110(15/22)^{t/30} \\
25 = 110(15/22)^{t/30} 
\frac{25}{110} = (15/22)^{t/30} \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}$

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What is the whole question

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2 years ago

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Answer:

t/8 + t/7 = 1

Step-by-step explanation:

Given in the question that,

time require for Jose to paint the house = 7 hours

time require for Brandon to paint the house = 8 hours

Suppose t means Full house painted.

<h3>To solve the question we have to figure out how much each of them can paint in ONE hour.</h3>

7 hours----t

1 hour ---- t/7

8 hours----t

1 hour ---- t/8

<h3>Equation</h3>

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