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Ksju [112]
3 years ago
11

When temperature is zero degree Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the correspondin

g Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C, the Celsius temperature, F(c).
Mathematics
1 answer:
7nadin3 [17]3 years ago
5 0

Answer:

\\ y = 1.8(x) + 32 or \\ y = \frac{9}{5}(x) + 32

or equivalently:

\\ F = 1.8(C) + 32 or \\ F = \frac{9}{5}(C) + 32

Step-by-step explanation:

To express the Fahrenheit temperature <em>as a linear function of the Celsius temperature</em>, F(c), we can proceed as follows.

We can use here <em>the two-point form</em> <em>equation</em> of a line:

\\ y-y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x-x_1) [1]

We are asked to express the <em>Fahrenheit temperature</em> as a function of <em>Celsius temperature</em>, so the independent variable, in this case, is <em>x</em> (Celsius temperature) and the dependent variable is <em>y</em> (Fahrenheit temperature).

When temperature is zero degree Celsius (\\x_1 = 0), the Fahrenheit temperature is 32 (\\y_1 = 32).

When the Celsius temperature is 100 (\\x_2 = 100), the corresponding Fahrenheit temperature is 212 (\\y_2 = 212).

Then, using [1], we have:

\\ y-32 = \frac{212 - 32}{100 - 0}(x-0)

\\ y-32 = \frac{180}{100}(x)

\\ y-32 = 1.8(x).

It could be also be written as:

\\ y-32 = \frac{18}{10}(x) = \\ y-32 = \frac{9}{5}(x), as it commonly appears in books.

Then <em>the Fahrenheit temperature express as a linear function of the Celsius temperature, F(c</em>) is ( solving the equation for <em>y </em>) :

\\ y = 1.8(x) + 32 or \\ y = \frac{9}{5}(x) + 32.

Or equivalently:

\\ F = 1.8(C) + 32 or \\ F = \frac{9}{5}(C) + 32

We can check this using the given values from the question:

For 0 Celsius degrees, the Fahrenheit temperature is:

\\ y = 1.8(0) + 32 = 32 Fahrenheit degrees.

For 100 Celsius degrees, the Fahrenheit temperature is:

\\ y = 1.8(100) + 32 = 180 + 32 = 212 Fahrenheit degrees.

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Let r= v ×2x ^ 2 + v × 1x +v0 , w= w ×2x ^ 2 + w × 1x +w0 . We have that\\v+w= (vO + wO) ^  x^ 2 +(vl^ × wl)^  x+ ( v 2^ × w2)• w+v= (wO + vO) ^x^ 2 +(wl^ × vl)x+ ( w 2^ ×v2)

Since the sum of actual numbers is commutative, we finish that v + w = w + v Therefore, belongings 5(a) is valid.

Property 5(b) isn't valid: we are able to introduce

a counter example. we could use z = 1 thenv = x ^ 2 w = x ^ 2 + 1\\(v + w) + z = (x ^ 2 + 2) + 1 = 3x ^ 2 + 1

v + (w + z) = x ^ 2 + (2x ^ 2 + 1) = x ^ 2 + 3

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Property f doesnt observe because of the switch of variables. for instance, if v = x ^ 2 1 × v=1^ × x ^ 2 = 1 #V

Properties 1,2, 5(a) and 5(c) are satisfied, the relaxation of the homes arent legitimate.

Step-with the aid of using-step explanation:

Note that each sum and scalar multiplication entails in replacing the order from that most important coefficient with the impartial time period earlier than doing the same old sum/scalar multiplication.

Property three does now no longer follow: Suppose that Property three is legitimate, shall we name v = a × x ^ 2 +bx +c the neuter of V. Since v is the neuter, then O have to be constant with the aid of using the neuted, consequently0 = O + v = (O × x ^ 2 + Ox + O) + (a × x ^ 2 + bx + c) = c × x ^ 2 + b ^ 2 + a

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Property 5(b) isn't valid: we are able to introduce

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Note that each expressions are same because of the distributive rule of actual numbers. Also, you could be aware that his assets holds due to the fact in each instances we 'switch variables twice.

· (1+2)^ * (x^ 2 +x)=3^ * (x ^ 2 + x) = 3x + 31^ * (x^ 2 +x)+2^ * (x ^ 2 + x) = (x + 1) + (2x + 2) = 3x ^ 2 + x ( ne 3x + 3 )(1^ * 2)^ * (x^ 2 +x)=2^ * (x ^ 2 + x) = 2x + 21^ * (2^ * (x ^ 2 + x) )=1^ ×* (2x+2)=2x^ 2 +2x( ne2x+2)

Read more about polynomials :

brainly.com/question/2833285

#SPJ4

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