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The picture below shows a thermometer with both Fahrenheit and Celsius temperatures. Temperature in degrees Fahrenheit, F, is a linear function of the temperature in degrees Celsius, C.
BOILING POINT OF WATER
By looking at the picture, identify two ordered pairs (C,F) that lie on the graph of this function. Find the slope between the two points = 140 and express F as a linear function of C.
100
CELSIUS
FREEZING POINT OF WATER
Using the function you found, determine the Celsius equivalent of 689 Fahrenheit. Show work and round your answer to two decimal places if needed. (If F = 68, C = 222)
There is a particular temperature, T, for which T degrees Fahrenheit = T degrees Celsius. Using your above function as well as the equation F = C, solve for that particular temperature. Show work.

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01:47

Solve each problem.Celsius to Falirenheit Formula Fahrenheit temperature $F$ is a linear function of Celsius temperature $C$. The ordered pair $(0,32)$ is an ordered pair of this function because $0^{\circ} \mathrm{C}$ is equivalent to $32^{\circ} \mathrm{F},$ the freezing point of water. The ordered pair $(100,212)$ is also an ordered pair of this function because $100^{\circ} \mathrm{C}$ is equivalent to $212^{\circ} \mathrm{F},$ the boiling point of water. Use the two given points and the point-slope formula to write $F$ as a function of $C .$ Find the Fahrenheit temperature of an oven at $150^{\circ} \mathrm{C}$

04:55

Let $x$ denote a temperature on the Celsius scale, and let $y$ denote the corresponding temperature on the Fahrenheit scale.(a) Find a linear function relating $x$ and $y,$ use the facts that $32^{\circ} \mathrm{F}$ corresponds to $0^{\circ} \mathrm{C}$ and $212^{\circ} \mathrm{F}$ corresponds.to $100^{\circ} \mathrm{C} .$ Write the function in the form $y=A x+B$(b) What Celsius temperature corresponds to $98.6^{\circ} \mathrm{F}$ ?(c) Find a number $z$ for which $z^{\prime} \mathrm{F}=z^{\prime} \mathrm{C}$.

01:07

The graph of Fahrenheit temperature, $^{\circ} \mathrm{F}$, as a function of Celsius temperature, $^{\circ} \mathrm{C}$, is a line. You know that $212^{\circ} \mathrm{F}$ and $100^{\circ} \mathrm{C}$ both represent the temperature at which water boils. Similarly, $32^{\circ} \mathrm{F}$ and $0^{\circ} \mathrm{C}$ both represent water’s freezing point.(a) What is the slope of the graph?(b) What is the equation of the line?(c) Use the equation to find what Fahrenheit temperature corresponds to $20^{\circ} \mathrm{C}$(d) What temperature is the same number of degrees in both Celsius and Fahrenheit?

01:08

There are two common systems for measuring temperature, Celsius and Fahrenheit. Water freezes at $0^{\circ}$ Celsius $\left(0^{\circ} \mathrm{C}\right)$ and $32^{\circ}$ Fahrenheit $\left(32^{\circ} \mathrm{F}\right) ;$ it boils at $100^{\circ} \mathrm{C}$ and $212^{\circ} \mathrm{F}$.(a) Assuming that the Celsius temperature $T_{C}$ and the Fahrenheit temperature $T_{F}$ are related by a linear equation, find the equation.(b) What is the slope of the line relating $T_{F}$ and $T_{C}$ if $T_{F}$ is plotted on the horizontal axis?(c) At what temperature is the Fahrenheit reading equal to the Celsius reading?(d) Normal body temperature is $98.6^{\circ} \mathrm{F}$. What is it in $^{\circ} \mathrm{C} ?$

Transcript

Okay. Um Okay we want to ordered pairs in this form. So looking at the thermometer I see 20 Celsius is 68. Fahrenheit and 10 Celsius is 50 F. And we want to um Find the slope between those. So we’ll do the change in Y 68 -50 Over change in X 20-10. So that is 18/10. If I divide by two that is 9/5. Mhm. Okay so the slope is 9/5 times are Celsius and we can see on the thermometer that the um zero Celsius is 32 F. So 32 would be our y intercept and that will equal Fahrenheit. Okay then using the function you found determine the Celsius equivalent of 68 F. So if nine fists c plus 32 is 68 Then we will subtract the 32 from both sides. 950 will equal 36. And then to get rid of the nine fish, you multiply those sides by five nines, cancel those guys out. So we do 36 times five and then divide by nine. And that…