4 years ago

HELP. I have tried and tried and I CAN NOT solve these equations!

Mathematics

1 answer:

umka21 [38]4 years ago

6 0

Here are the solutions. See the attachment. I've color-coded for clarity.

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Water boils at 100LaTeX: ^\circ∘ Celsius or 212LaTeX: ^\circ∘ Fahrenheit. Water freezes at 0LaTeX: ^\circ∘ Celsius or 32LaTeX: ^

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Answer:

$F = 1.8C +32$

$F = 86^{\circ}F$

Step-by-step explanation:

Represent Celsius with C and Fahrenheit with F.

So, we have:

$(C_1,F_1) = (100,212)$

$(C_2,F_2) = (0,32)$

Solving (a): Represent as a linear equation

First, we calculate the slope:

$m = \frac{F_2 - F_1}{C_2 - C_1}$

This gives:

$m = \frac{32 - 212}{0 - 100}$

$m = \frac{-180}{- 100}$

$m = 1.8$

The linear equation is then calculated using:

$F - F_1 = m(C -C_1)$

Where

$(C_1,F_1) = (100,212)$

$m = 1.8$

$F - 212 = 1.8(C -100)$

$F - 212 = 1.8C -180$

Make F the subject

$F = 1.8C -180+212$

$F = 1.8C +32$

Solving (b): The value of F when C = 30

Substitute 30 for C in $F = 1.8C +32$

$F = 1.8 * 30 + 32$

$F = 54 + 32$

$F = 86$

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