Knowing their linear relationship, find the expression of {∘}C in terms of {∘}F using the point-slope form af analytic geomtry. With boiling of 100∘C=212∘F and the freezing point of 0∘C=32∘F

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Karen Robbins · Accepted Answer

Step 1  If we have 2 points (x0, y0) and (x1, y1) then the point-slope form is y−y1=m(x−x1), here, m is the slope given by:  m=y1−y0x1−x0  Hence, the point-slope form becomes y−y1=(y1−y0x1−x0)(x−x1)  Now, we have to find the relation between Fahrenheit and Celcius.   Let Fahrenheit be represented by F and Celcius by C.  Now, we are given that 100∘C=212∘F and 0∘C=32∘F  This means that when the temperature is 100∘C then it is same as 212∘F, and when the temperature is 0∘C it is same as 32∘F  Therefore, let:  C0=0∘C and F0=32∘F  C1=100∘C and F1=212∘F  Step 2  Therefore, we receive two points  (C0, F0)=(0, 32)  (C1, F1)=(100, 212)  Now, we find the point-slope form:  F−F1=(F1−F0C1−C0)(C−C1)  ⇒F−212=(212−32100−0)(C−100)  ⇒F−212=(180100)(C−100)  ⇒F−212=(95)(C−100)  ⇒F−212=95C−(95)100  ⇒F−212=95C−9×20  ⇒F−212=95C−180  ⇒F=95C+32

Jenny Sheppard · Accepted Answer

Step 1  The relatijn between the celsium and Fanrenhcit scals  As the temperature risce on the celsius scale the equivalent temperature of fahrenhcit waild also be high  As the temperature falls on the celsius scale, the temp equal to fahrenhciet would also be low-  freezing point of water at {∘}C=0∘C  and boiling point of untered {∘}C=100∘C  freezing point of water at Fahrenhcit =32∘F  Boiling point  According to thed  C−L×F×PU×F×P−L×F×P=F−L×F×PU×F×P−L×F×P  L×F×P=Lower Fixed point  U×F×P=Upper fixed point  C−O100−0=F−32212−32  C100=F−32180  C5=(F−32)9  C=59(F−32)  C=59F−5×329  c=59F−1609

nick1337 · Accepted Answer

Step 1 If C is a function, then we may represent the points as (F, C) (C−C1)=m(F−F1) (32,0) (212.100) (C−0)=100−0212−32(F−32) C=100180(F−32) C=59(F−32)

Knowing their linear relationship, find the expression of ^{\circ}C in

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