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Mama L [17]
1 year ago
11

Find the coordinates of point z such that the ratio of fz to zd is 5;3

Mathematics
1 answer:
Vilka [71]1 year ago
8 0

The point such that the coordinate is 5;3 is (14, 0)

<h3>Midpoint of coordinates using ratio</h3>

The formula for finding the midpoint of a line in the ratio m:n is expressed as:

M(x, y) = {(mx₁+nx₂)/2, (my₁+ny₂)/2,}

Given the coordinate of G and D on the line as G(5, 0) and D(1,0)

Since there is no y-axis, hence;

x = 5(5)+1(3)/2

x = 25+3/2

x = 28/2

x =14

Hence the point such that the ratio is 5;3 is (14, 0)

Learn more on midpoint of a line here: brainly.com/question/5566419

#SPJ1

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Which ratio is greater? 4:3 or 7:9. (Show necessary steps and calculations for your answer)
scoundrel [369]

Answer:

4 : 3 > 7 : 9

Step-by-step explanation:

In order to compare two ratios or fractions, we make their denominators equal and then equate their numerators.

4 : 3 =  \frac{4}{3}  =  \frac{4 \times 3}{3 \times 3}  =  \frac{12}{9}  \\  \\ 7 : 9 =  \frac{7}{9} \\  \\  \because \: 12 > 7 \\  \\  \therefore \:  \frac{4}{3} > \frac{7}{9} \\  \\  \implies \: 4 : 3  > 7 : 9

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3 years ago
What is the median of this set of data one, two, five, six, nine
Alekssandra [29.7K]

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5

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Viefleur [7K]

Let point A(x,y) be the point on the x-axis. All points that lie on the x-axis have y-coordinate equal to 0, then A(x,y).

Find the distance from point A to point (12,-5):

d=\sqrt{(x-12)^2+(0-(-5))^2}=\sqrt{(x-12)^2+25}.

This distance is equal to 13, then

\sqrt{(x-12)^2+25}=13,\\ \\(x-12)^2=169-25,\\ \\(x-12)^2=144,\\ \\x-12=12 \text{ or } x-12=-12,\\ \\ x=24 \text{ or }x=0.

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6 0
3 years ago
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