Answer:
D, 5/13
Step-by-step explanation:
The cosine of an angle is the adjacent side over the hypotenuse. In this case, that is 5/13. Hope this helps!
To solve this problem you must apply the proccedure shown below:
1. You have the following standard form for the hyperbola given in the problem above: <span> (x-2)^2/36 - (y+1)^2/64=1
</span> 2. Therefore, you can calculate the lengtn of the transverse axis as following:
Length of transverse axis=2a
a^2=36
a=√36
a=6
Length of transverse axis=2(6)=12
Then, the answer is:12
Answer: a: 4:5, b: 5:4, c: 5:9
Step-by-step explanation:
D is halfway between A and B
so the coordinates of D are (2,2)
E is halfway between A and C so the coordinates of E are (-1,1)
now you need to find the gradient/slope of DE and BC using the formula:

<h3>
<u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>D</u><u>E</u><u>:</u><u> </u></h3>
SUB IN COORDINATES OF D AND E

therefore the gradient of DE is 1/3.
<h3>
<u>G</u><u>r</u><u>a</u><u>d</u><u>i</u><u>e</u><u>n</u><u>t</u><u> </u><u>o</u><u>f</u><u> </u><u>B</u><u>C</u><u>:</u></h3>
<em>S</em><em>U</em><em>B</em><em> </em><em>I</em><em>N</em><em> </em><em>C</em><em>O</em><em>O</em><em>R</em><em>D</em><em>I</em><em>N</em><em>A</em><em>T</em><em>E</em><em>S</em><em> </em><em>O</em><em>F</em><em> </em><em>B</em><em> </em><em>A</em><em>N</em><em>D</em><em> </em><em>C</em>
<em>
</em>
therefore the gradient of BC is -2/-6 which simplifies to 1/3.
<h3>
therefore, BC and DE are parallel as they both have a gradient/slope of 1/3 and parallel lines have the same gradient</h3>
Answer: It’s A y > x^2-4
Y < x^2+3
Step-by-step explanation: