School 2x still sucks

Mathematics

2 answers:

Alexeev081 [22]3 years ago

6 0

This question was asked almost a year ago, there is a very good chance that they're not even in school anymore, it's crazy to think about. A year ago, I was hesitant on wearing a mask for the first couple of times and was starting to get used to the changes that were soon to come, while this guy was here, begging people for answers to a question so he could pass highschool. Now, he's probably living his best life right now and doesn't even know what I'm typing right now, either that, or he's working at McDonald's. Honestly though, time flies by so fast though, it really is crazy.

garik1379 [7]3 years ago

5 0

Answer: 1

Explanation:

$1\frac{3}{5} = \frac{8}{5}$

$\frac{8}{5} * \frac{5}{8} = \frac{40}{40}$

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Pleasee help, thank you so much

dezoksy [38]

Answer:

-2

Step-by-step explanation:

Use the slope formula when given two points

Slope formula = $\frac{y_2-y_1}{x_2-x_1}$

( the x and y values are determined by the given points )

Given the points ( 0 , 5 ) and ( 1 , 3 )

Plug in the x and y values of the given points into the formula

( remember y values on top x values on bottom )

$slope=\frac{5-3}{0-1} \\5-3=2\\0-1=-1\\slope=\frac{2}{-1} =-2$

Hence, the answer is D

5 0

3 years ago

1. A boat is 18 ft away from a point perpendicular to the shoreline. A person stands at a point down the shoreline so that a 68°

slega [8]

Answer:

19.4 feet

Step-by-step explanation:

When calculating distances using perpendicular angles, we use trig functions such as Sin, Cos, and Tan.

Sine is defined in a perpendicular triangle as the ratio of the opposite side to the angle over the hypotenuse.
Cosine is defined in a perpendicular triangle as the ratio of the adjacent side to the angle over the hypotenuse.
Tangent is defined in a perpendicular triangle as the ratio of the opposite side over the adjacent side.

Because our angle and know side value are across from each other and we need to know the hypotenuse, we chose to use Sine.

We set up the equation $sin (68) = \frac{18}{h}$ .

We isolate h by multiplying h across the equal sign and dividing sin (68) across as well. $sin (68) = \frac{18}{h}\\hsin (68) = \frac{18}{h}*h\\hsin (68) = 18\\\frac{hsin (68)}{sin (68)} =\frac{18}{sin (68)} \\$