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timama [110]
3 years ago
9

The midpoint of line segment AB is (1,2). if the coordinates of A are (1,0) what are the coordinates of B

Mathematics
1 answer:
seraphim [82]3 years ago
7 0

The required coordinate of B will be at (1, 4)

When a point bisects a line, it cuts the line into two equal parts and the point on that line is known as its midpoint. The expression for calculating the midpoint of a line is expressed as;

M(X, Y)= [{ \frac{x_2+x_1}{2} , \frac{y_2+y_1}{2} ] where:

X =\frac{x_2+x_1}{2} \\Y=\frac{y_2+y_1}{2}

From the question, we are given the following coordinate points

X=1\\x_1=1\\

Get x₂;

X =\frac{x_2+x_1}{2}\\1 =\frac{x_2+1}{2}\\2=x_2+1\\x_2=2-1\\x_2=1\\

Get y₂;

Y =\frac{y_2+y_1}{2}\\2 =\frac{y_2+0}{2}\\2 \times 2=y_2+0\\4=y_2+0\\y_2=4\\

Hence the coordinate of B will be (1, 4)

Learn more about the midpoint here: brainly.com/question/18315903

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The equation C = 5 9 F − 160 9 gives the relation between temperature readings in Celsius and Fahrenheit.
Pavel [41]

Step-by-step explanation:

The question is wrong. The correct equation is :

C=\frac{5}{9}F-\frac{160}{9}

We know that the equation gives the relation  between temperature readings in Celsius and Fahrenheit.

Therefore, giving that we know the value in Fahrenheit ''F'' we can find the reading in Celsius ''C''. This define a function C(F) that depends of the variable ''F''.

So for the incise (a) we answer Yes, C is a function of F.

For (b) we need to find the mathematical domain of this function. Giving that we haven't got any mathematical restriction, the mathematical domain of the function are all real numbers.

Dom (C) = ( - ∞ , + ∞)

For (c) we know that the water in liquid state and at normal atmospheric pressure exists between 0 and 100 Celsius.

Therefore the range will be

Rang (C) = (0,100)

Now, we need to find the domain for this range. We do this by equaliting and finding the value for the variable ''F'' :

For C = 0 :

0=\frac{5}{9}F-\frac{160}{9} ⇒ F=32

And for C = 100 :

100=\frac{5}{9}F-\frac{160}{9} ⇒ F=212

Therefore, the domain as relating temperatures of water in its liquid state is

Dom (C) = (32,212)

For (d) we only need to replace in the equation by F=71 and find the value of C ⇒

C=\frac{5}{9}F-\frac{160}{9} ⇒

C=(\frac{5}{9})(71)-\frac{160}{9}

C=\frac{65}{3} ≅ 21.67

C(71) = 21.67 °C

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