Answer:
It is a many-to-one relation
Step-by-step explanation:
Given
See attachment for relation
Required
What type of function is it?
The relation can be represented as:
![\left[\begin{array}{c}x\\ \\-5\\-2\\1\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5C%20%5C%5C-5%5C%5C-2%5C%5C1%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}y\\ \\10\\11\\4\\10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dy%5C%5C%20%5C%5C10%5C%5C11%5C%5C4%5C%5C10%5Cend%7Barray%7D%5Cright%5D)
Where
and 
Notice that the range has an occurrence of 10 (twice)
i.e.
and 
In function and relations, when two different values in the domain point to the same value in the range implies that, <em>the relation is many to one.</em>
Answer:
i think 64 for your answer
Step-by-step explanation:
Answer:
yes I believe so. best of luck
Answer:
(5,9)
Step-by-step explanation:
3y=4x+7
5y=4x+25
Subtract the two equations to eliminate x
5y=4x+25
-( 3y=4x+7)
------------------
2y = 0x +18
Divide by 2
2y/2 = 18/2
y =9
Now find x
3y = 4x+7
3*9 = 4x+7
27 = 4x+7
Subtract 7
20 = 4x
Divide 4
20/4 = 4x/4
5 =x
(5,9)
5. Given the equation y = (1/4) cos[(2pi/3)*theta]:
5a. For the general equation y = a cos(bx), the period is given by 2pi/b. In this equation, b = 2pi/3, so this means that 2pi/b = 2pi/(2pi/3) = 3. Therefore, the period of this equation is 3, and the cosine wave repeats itself every 3 x-units.
5b. For the general equation y = a cos(bx), the amplitude is given by a. Therefore the amplitude is a = 1/4, and this means that the cosine wave's range is from -1/4 to 1/4 for all values of x.
5c. The equation of the midline is y = 0. This represents the average value over the wave. This is determined by adding the highest and lowest values of the range and taking the average, in this case, 1/4 + (-1/4) = 0, and 0 / 2 = 0. Another way to do this is using the general equation y = a cos(bx) + c, where the midline's equation is y = c. In this case, there is no value of c in the given, implying that c = 0, and the midline is y = 0.
6. Let the horizontal distance be x. Then tan42 = h/x, and h = x tan42. Then using the Pythagorean theorem: 3280^2 = h^2 + x^2
3280^2 = x^2 (tan42)^2 + x^2
3280^2 = x^2 [(tan42)^2 + 1]
x = 2437.52
Therefore, h = x tan42 = 2,194.75 ft.
