Mathematics: WILL MARK BRAINLIEST HELP ASAP!! A caterpillar crawling in a straight line across a coordinate plane starts at point (—3, —4) and ends at point (25, 38). Counting the start point and end point, how many points with integer coordinates does the caterpillar touch?

We can use 3 facts:So, to find our x and our y we can do:So, now that we know our x and our y we can find our r.You ll notice, though, that there were two solutions for , which means we have a positive and a negative for our x. So:...

WILL MARK BRAINLIEST HELP ASAP!! A caterpillar crawling

Ответ:

Sashakimora2020

10.04.2022

Mathematics

Step-by-step explanation:

The slope = $\frac{change in y}{change in x} =\frac{38+4}{25+3} =\frac{42}{28} =\frac{3}{2}$

This means that when the caterpiller moves 2 units to the right, it moves 3 units up. Since 42/3 = 28/2 = 14. this happens 14 times from (-3, -4) to (25, 28). Counting the starting point (-3, -4) we get 14 + 1 = 15 points with integer coordinates.

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Ответ:

jungkookie001

28.06.2022

Mathematics

We can use 3 facts:

$\tan{\theta}=\frac{y}{x}$

$\sec{\theta}=\frac{r}{x}$

x^2+y^2=r^2

So, to find our x and our y we can do:

$\tan^2{\theta}=\frac{3}{8} \implies \\ \tan{\theta}=\pm{\sqrt{\frac{3}{8}}=\pm\frac{\sqrt{3}}{\sqrt{8}}$ So, now that we know our x and our y we can find our r.

$(\sqrt{3})^2+(\sqrt{8})^2=11 \implies r=\sqrt{11}$

You'll notice, though, that there were two solutions for $\tan{\theta}$ , which means we have a positive and a negative for our x. So:

$\sec{\theta}=\frac{\sqrt{11}}{\sqrt{8}} \\ \sec{\theta}=-\frac{\sqrt{11}}{\sqrt{8}}}\\ \text{Rationalizing both of them:}\\ \sec{\theta}=\frac{\sqrt{88}}{8} \\ \sec{\theta}=-\frac{\sqrt{88}}{8}$

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WILL MARK BRAINLIEST HELP ASAP!! A caterpillar crawling in a straight line across a coordinate plane starts at point (—3, —4) and ends at point (25, 38). Counting the start point and end point, how many points with integer coordinates does the caterpillar touch?

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