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3 years ago

Find the coordinates of the point three tenths of the way from A(-4, -8) to B(11, 7)

Mathematics

1 answer:

Leviafan [203]3 years ago

4 0

let's first off take a peek at those values.

let's say the point with those coordinates is point C, so C is 3/10 of the way from A to B.

meaning, we take the segment AB and cut it in 10 equal pieces, AC takes 3 pieces, and CB takes 7 pieces, namely AC and CB are at a 3:7 ratio.

$\bf ~~~~~~~~~~~~\textit{internal division of a line segment}
\\\\\\
A(-4,-8)\qquad B(11,7)\qquad
\qquad \stackrel{\textit{ratio from A to B}}{3:7}
\\\\\\
\cfrac{A\underline{C}}{\underline{C} B} = \cfrac{3}{7}\implies \cfrac{A}{B} = \cfrac{3}{7}\implies 7A=3B\implies 7(-4,-8)=3(11,7)\\\\[-0.35em]
~\dotfill\\\\
C=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em]
~\dotfill$

$\bf C=\left(\cfrac{(7\cdot -4)+(3\cdot 11)}{3+7}\quad ,\quad \cfrac{(7\cdot -8)+(3\cdot 7)}{3+7}\right)
\\\\\\
C=\left( \cfrac{-28+33}{10}~~,~~\cfrac{-56+21}{10} \right)\implies C=\left( \cfrac{5}{10}~~,~~\cfrac{-35}{10} \right)
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hfill C=\left( \frac{1}{2}~,~-\frac{7}{2} \right)~\hfill$

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3 years ago

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Step-by-step explanation:

The slope intercept form is:

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Slope intercept form is useful for finding the slope and y intercept of a line, hence why it is called "slope intercept" because it is easy to see the slope and the y intercept of a line in this form. Since the slope denoted by m, and y intercept denoted by b are clearly given.

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3 years ago

Does anybody know how to do time with exponential decay

My name is Ann [436]

From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

$b_n=0.12\times b_{n-1}$

Why does this work? Initially, you start with 500 mL of air that you breathe in, so $b_1=500\text{ mL}$ . After the second breath, you have 12% of the original air left in your lungs, or $b_2=0.12\timesb_1=0.12\times500=60\text{ mL}$ . After the third breath, you have $b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}$ , and so on.

You can find the amount of original air left in your lungs after $n$ breaths by solving for $b_n$ explicitly. This isn't too hard:

$b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots$

and so on. The pattern is such that you arrive at

$b_n=0.12^{n-1}b_1$

and so the amount of air remaining after $50$ breaths is

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which is a very small number close to zero.

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