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Mandarinka [93]
1 year ago
12

In 1911, the temperature in Rapid City,

Mathematics
1 answer:
Sliva [168]1 year ago
8 0

The 3.1 °F/min rate of change of the temperature and 15 minutes change duration gives the change in temperature as 46.5 °F

<h3>How can the change in temperature be found from the rate of change?</h3>

The rate at which the temperature changed = 3.1 °F/min

The duration of the change in temperature = 15 minutes

The relationship between the change in temperature, the rate of change in temperature and the time can be presented as follows;

3.1  ^{\circ} F/min =  \frac{ \delta T }{\delta t}  =  \frac{ \Delta T }{ \Delta t}

Where;

∆T = The required change in temperature

∆t = The duration of the change = 15 minutes

Which gives;

∆T = 3.1°F/min × 15 minutes = 46.5 °F

  • The change in temperature, ∆T = 46.5 °F

Learn more about the rate of change of a variable here:

brainly.com/question/10208814

#SPJ1

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QUESTION 11.1
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The group paid $ 5250 at first city and $ 6250 at second city

<u>Solution:</u>

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the  second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the  tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

<em><u>Therefore, a equation is framed as:</u></em>

8.5 % of x + 5.5 % of y = 790

\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790

0.085x + 0.055y = 790 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Substitute eqn 1 in eqn 2</u></em>

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

<h3>x = 5250</h3>

<em><u>Substitute x = 5250 in eqn 1</u></em>

y = 5250 + 1000

<h3>y = 6250</h3>

Thus the group paid $ 5250 at first city and $ 6250 at second city

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3 years ago
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