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usre123
According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A)9
B)15
C)47
D) 48 3/5
E) 59

\(F_i = \frac{9}{5}* (C_i) +32\)

\(F_i+27= \frac{9}{5}* (C_f) +32\)

On Subtracting, we get\(27 = \frac{9}{5}* (C_f-C_i) \to (C_f-C_i) = 3*5 = 15\)

B.
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Bunuel
usre123
According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A)9
B)15
C)47
D) 48 3/5
E) 59

You can plug in values.

C = 5/9*(F-32)

F=32 --> C=0;
F=32+27=59 --> C=5/9*27=15.

Increase = 15 degrees.

Answer: B.

I dont understand why you only multiplied 5/9 by 27. what about F-32?
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usre123
Bunuel
usre123
According to the formula F=9/5 (C) +32, if the temperature in degrees Farenheit (F) increases by 27, by how much does the temperature in degrees Celsius (C) increase?

A)9
B)15
C)47
D) 48 3/5
E) 59

You can plug in values.

C = 5/9*(F-32)

F=32 --> C=0;
F=32+27=59 --> C=5/9*27=15.

Increase = 15 degrees.

Answer: B.

I dont understand why you only multiplied 5/9 by 27. what about F-32?

If F=59, then \(C = \frac{5}{9}*(F-32) = \frac{5}{9}*(59-32) = \frac{5}{9}*27 = 15\).

Does this make sense?
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Could someone please explain what happens to the F=9/5 (C) +32? In all of the provided solutions, no one does anything with this, but I'm confused as to why that is.
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VeritasPrepKarishma
skimmingit
Could someone please explain what happens to the F=9/5 (C) +32? In all of the provided solutions, no one does anything with this, but I'm confused as to why that is.


32 is constant. It doesn't bring about the change. The change is defined by the relation between F and C only. 32 just gets cancelled out.

F1 = (9/5)*C1 + 32 ......(I)
F2 = (9/5)*C2 + 32 ......(II)

(II) - (I)

F2 - F1 = (9/5)*(C2 - C1) + 32 - 32

F2 - F1 = (9/5)*(C2 - C1)

27 = (9/5)*(C2 - C1)

(C2 - C1) = 15

hi mam

since the change is defined by the relation between F and C only, we can say directly that the "change in F" is equal to 9/5 times the "change in C". That's it...

thanks mam .. :thumbup:
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Given expression is in terms of F but we need the value of C, hence we will express the expression in terms of C.

F = \(\frac{9}{5}\) * C + 32

At C = 0, F = 32

Increased value: F = 32 + 27 = 59

=> C = \(\frac{5}{9}\) * ( F - 32)

=> C = \(\frac{5}{9}\) * ( 59 - 32)

=> C = \(\frac{5}{9}\) * 27

=> C = 15

Answer B
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