Answer:
a = 
Step-by-step explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one.:
<em>Suppose ABC is a right triangle with sides a, b, and c and right angle at C. Find the unknown side length using the Pythagorean theorem and then find the values of the six trigonometric functions for angle B. when b=3 and c=4</em>.
My answer:
We will Pythagoras theorem, which states that the sum of squares of two legs of a right triangle is equal to the square of the hypotenuse of right triangle. Because the question says that ABC is a right triangle.

Given that: b=3 and c=4

so a =
We know that tangent relates opposite side of a right triangle with adjacent side.

Please have a look at the attached photos.
<span> we know the length of the cable is 9m.
That means the magnitude of </span><span><span>r<span><span><span>AB</span></span><span></span></span></span>=9</span><span>m.
The unit vector, denoted u, is each of </span><span>r<span><span><span>AB</span></span><span></span></span></span><span> divided by the magnitude.
</span>u=<span>(<span><span>x/9</span><span></span></span>i−<span>y/9<span></span></span>j−<span>z/9<span></span></span>k<span>)
</span></span><span>We can also figure out the unit vector of F.
</span>u=(350i - 250j - 450k)/√(350² +(-250)² +(-450)²)
u=0.562i−0.401j−0.723<span>k
</span>
<span>Force F is directed from point A to B, then both unit vectors must be equal.
Therefore
</span>(x/9i−y/9j−z/9k)=0.562i−0.401j−0.723k
<span>We can now solve for each term
x/9=0.562----- > x=5.058 m
-y/9=-0.401--------- > y=3.609 m
-z/9=-0.723------- > z=6.507 m
The answer is
the coordinates of point a
(5.058,3.609,6.507)
</span>
What is the mode of this data set? {8, 11, 20, 10, 2, 17, 15, 5, 16, 15, 25, 6}
lidiya [134]
The mode is what appears the most so 15 is the answer because it appears the most.
If you looking for the amount of beans
25 = 16.75 + 1.50×b
25-16.75 = 1.50b
8.25 = 1.50b
8.25/1.50 = b
b = 5.5
1. $325 2. $315,000 3.$2,600 4.$6,00 5.i don’t know