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Lostsunrise [7]
2 years ago
7

Using algebra, find the point at which the line k(x) = 5x - 1 intersects with the line h(x) = -3x - 1 ?

Mathematics
1 answer:
vichka [17]2 years ago
8 0

The point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Given: k(x) = 5x - 1, h(x) = -3x - 1

We need to find the point(if any) at which these two lines k and h meets.

To find point of intersection(if any), we need to set the functions equal as at the point of intersection the (x, y) value will be same for both of the lines.

Therefore, k(x) = h(x)

=> 5x - 1 = -3x - 1

=> 8x = 0

=> x = 0

k(x=0) = 5 * 0  - 1 = -1

Hence the point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Know more about "point of intersection" problems here: brainly.com/question/16929168

#SPJ1

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A fudge recipe calls for 2 1/3 cups of flour for every 3/4 cups of cocoa powder. Could you increase the recipe proportionately b
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same as before, is the proportion of one, the same as the other?  let's do the same here without much fuss.


\bf \stackrel{mixed}{2\frac{1}{3}}\implies \cfrac{2\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{7}{3}}
~\hfill
\stackrel{mixed}{4\frac{2}{3}}\implies \cfrac{4\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{14}{3}}


\bf \stackrel{mixed}{1\frac{1}{2}}\implies \cfrac{1\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{3}{2}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\cfrac{~~2\frac{1}{3}~~}{\frac{3}{4}}=\cfrac{~~4\frac{2}{3}~~}{1\frac{1}{2}}\implies \cfrac{~~\frac{7}{3}~~}{\frac{3}{4}}=\cfrac{~~\frac{14}{3}~~}{\frac{3}{2}}\implies \cfrac{7}{3}\cdot \cfrac{4}{3}=\cfrac{14}{3}\cdot \cfrac{2}{3}\implies \cfrac{28}{9}=\cfrac{28}{9}~~\textit{\Large \checkmark}

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