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2 years ago

5. Is the side that is 100 meters long the hypotenuse, opposite or adjacent side? *

Mathematics

1 answer:

Stolb23 [73]2 years ago

7 0

Answer:

adjacent

Step-by-step explanation:

It is the leg that touches the 18 degree angle

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Which graph represents the polar curve r = 4cos(30)?

stepan [7]

The equation of the polar curve $r = 4\cos(3\theta)$ is an even cosine function.

<h3>Polar curves</h3>

In geometry, polar curves are curves that are represented by the polar coordinates such that they are drawn around a fixed point

The equation of the polar curve is given as:

$r = 4\cos(3\theta)$

To graph the polar curve, we make use of the following representations:

r represents the y-axis
$\theta$ represents the x-axis

Using the above representation, we can graph the polar curve

See attachment for the graph that represents $r = 4\cos(3\theta)$

Read more about polar curves at:

brainly.com/question/4618020

8 0

2 years ago

A company that manufactures toy trains is making a new boxcar. To fit on the track, the smallest dimension of the toy boxcar mus

pochemuha

Answer:

Length = 12 inches, height = 3 inches

Step-by-step explanation:

First, you have to set up a proportion. 2/7, x/42.

Then, you cross multiply and get the setup of 84=7x

Finally, you divide 84 by 7 and get 12.

12 is the length, and there is no other answer that has 12 in the height. You could just do the same proportion to get 3, though.

8 0

3 years ago

Read 2 more answers

2 over 5 (x − 4) = 2x.

Sav [38]

2/5 (x-4)=2x answer is x= -1

5 0

3 years ago

Write an algebraic expression for the phrase the sum of g and 3

Paraphin [41]

The sum of g and 3.

The sum of two values is added together.

g+3

You would add g to 3 since it is their sum the statement is asking for.

I hope this helps!
~kaikers

3 0

3 years ago

Let v1 =(-6,4) and v2=(-3,6) compute the following what i sthe angle between v1 and v2

Agata [3.3K]

$\bf ~~~~~~~~~~~~\textit{angle between two vectors }
\\\\
cos(\theta)=\cfrac{\stackrel{\textit{dot product}}{u \cdot v}}{\stackrel{\textit{magnitude product}}{||u||~||v||}} \implies 
\measuredangle \theta = cos^{-1}\left(\cfrac{u \cdot v}{||u||~||v||}\right)\\\\
-------------------------------$

$\bf \begin{cases}
v1=\ \textless \ -6,4\ \textgreater \ \\
v2=\ \textless \ -3,6\ \textgreater \ \\
------------\\
v1\cdot v2=(-6\cdot -3)+(4\cdot 6)\\
\qquad \qquad 42\\
||v1||=\sqrt{(-6)^2+4^2}\\
\qquad \sqrt{52}\\
||v2||=\sqrt{(-3)^2+6^2}\\
\qquad \sqrt{45}
\end{cases}\implies \measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{52}\cdot \sqrt{45}} \right)
\\\\\\
\measuredangle \theta =cos^{-1}\left( \cfrac{42}{\sqrt{2340}} \right)\implies \measuredangle \theta \approx 29.74488129694^o$

8 0

3 years ago