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Sidana [21]
3 years ago
15

8. In the 21st century, people measure length in feet and meters. At various points in history, people measured length in hands,

cubits, and paces. There are 9 hands in 2 cubits. There are 5 cubits in 3 paces.
11. Write an equation to express the relationship between hands, h, and paces, p.
Mathematics
1 answer:
velikii [3]3 years ago
7 0
The answer is 4.5h=c
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Find the equation of a line perpendicular to y - 12 = 2x – 8 that passes through the point (2, 3). (answer in slope-intercept fo
Alla [95]

Answer:

\displaystyle y=-\frac{1}{2}x+4

Step-by-step explanation:

<u>Equation of a Line</u>

We can find the equation of a line by using two sets of data. It can be a pair of ordered pairs, or the slope and a point, or the slope and the y-intercept, or many other combinations of appropriate data.

We are given a line

y - 12 = 2x -8

And are required to find a line perpendicular to that line. Let's find the slope of the given line. Solving for y

y = 2x +4

The coefficient of the x is the slope

m=2

The slope of the perpendicular line is the negative reciprocal of m, thus

\displaystyle m'=-\frac{1}{2}

We know the second line passes through (2,3). That is enough information to find the second equation:

y-y_o=m'(x-x_o)

\displaystyle y-3=-\frac{1}{2}(x-2)

Operating

\displaystyle y=-\frac{1}{2}(x-2)+3

Simplifying

\displaystyle y=-\frac{1}{2}x+4

That is the equation in slope-intercept form. Intercept: y=4

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3 years ago
Use the protractor to measure<br> these angles<br> M m m Check
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2 years ago
Find the solution set.<br> x^2 – 5x = 0
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Answer:

x=0  and x=5

Step-by-step explanation:

x^2 – 5x = 0

Factor out an x

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Using the zero product property

x=0 and x-5 =0

Solving

x=0 x-5+5 = 0+5

x=0  and x=5

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If (m, 2) is on a circle with center (−3, 2) and radius 8, what is a value of m?
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\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;&({{ -3}}\quad ,&{{ 2}})\quad &#10;%  (c,d)&#10;&({{ m}}\quad ,&{{ 2}})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf \textit{we know the distance from m,2 and -3,2(center) is the radius}&#10;\\\\\\&#10;d=8\implies 8=\sqrt{[m-(-3)]^2+[2-2]^2}&#10;\\\\\\&#10;8=\sqrt{(m+3)^2+(0)^2}\implies 8=\sqrt{(m+3)^2}\implies 8^2=(m+3)^2&#10;\\\\\\&#10;64=(m+3)^2\implies \sqrt{64}=\sqrt{(m+3)^2}\implies 8=m+3&#10;\\\\\\&#10;8-3=m
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