Answer:
11.6in
Step-by-step explanation:
The standard form of the equation of a circle is 
.
Solution:
Center of the circle = (6, 4)
Radius of the circle = 
<u>To find the equation of the circle:</u>
General formula for the equation of a circle
where (h, k) is the center of the circle and r is the radius of the circle.
Here, h = 6, k = 4 and r =
Hence the standard form of the equation of a circle is .
Answer: 21
Step-by-step explanation: This is a combination of 7 objects taken 5 at a time
nCr = n! / r! * (n - r)!
7C5 = 7! / (5! (7-5)!)
= 7! /(5! 2!)
= (7*6*5*4*3*2*1)
-----------------------
5*4*3*2*1 * 2*1
Canceling
= 7*6/ (2*1)
= 21
Since both y squared and 6y have a common factor of y, can put that to the side and divide both the y squared and 6y to get the second step of the work shown above. Separate both the y on the outside of the parentheses and the numbers within the parentheses and set them both equal to 0 as shown in step 3 of the work shown above. Finally solve for y as you would normally. Those two number in step 4 should be your final answer.
1. B & D are TRUE
3. C. No. It does not repeat on regular intervals
5. C.x|x=k/4 pi for every integer k
1. f(x)=cot x
First, plot a graph of cot(x) using whatever tools you desire. If you do that, you'll see several properties of the graph that make most of the answers rather obvious. Let's see what options are true.
A. It is an even function (FALSE)
A function is even iff f(-f) = f(x) for all x. So let's test that.
cot(pi/4) = 1
cot(-pi/4) = -1
1 is not equal to -1, so the function is NOT even.
B. Has an asymptote at x=0 (TRUE)
cot(x) is defined as cos(x)/sin(x). At x=0, cos(x) will be 1. But sin(x) will be
approaching 0 as x approaches 0. But if x is approaching 0 from negative values, sin(x) will be negative. And is x is approaching 0 from positive values, sin(x) will be positive. So cos(x)/sin(x) approaches either positive, or negative infinity depending upon the direction with which x is approaching 0. And that's pretty much the definition of an asymptote.
C.Has a zero at x=0 (FALSE)
Since option B shows that we have an asymptote at x=0, we can't have a zero at x=0 as well. So this is false.
d.has a period of pi. (TRUE)
This is obvious from inspection of the graph (you did plot the graph didn't
you?).
3.Look at the graph of f(x)=sin 1/x. is the graph periodic? why or why not?
Once again, plot the graph and look for yourself. If you actually plot the graph, the answer becomes obvious.
a. yes it is periodic. the value of f(x) varies between -1 and 1 repeatedly
* Just because it keeps the same range doesn't make it periodic. It also has to have a regular period and this function does not. So this is a bad choice.
b. yes it is periodic. its a transformation of sin (x)
* This function doesn't transform sin(X).And just because you have a transformation of sin(x) doesn't mean the function is periodic. So this is a bad choice.
c. no. it does not repeat on regular intervals
* EXACTLY. The function repeats, but doesn't repeat at a regular interval. So this is the correct choice.
d. no. it is a transformation of 1/x
* Just because a function is a transformation of another function doesn't make the transformed function periodic, or non-periodic. So this is a bad choice.
5. what set describes the zeroes of the function f(x)=6pi sin (4x) shown in the graph.
The zeros for the function will occur at each point the sin function returns 0 which is every integer multiple of pi. Now let's look at the available options and see which ones give every integer multiple of pi.
a.x|x=k pi for every integer k
* Yes, this will give only zeros. But it won't give every zero. For instance, the value pi can not be achieved. So this is a bad choice.
b. x|x=4kpi for every integer k
* Same problem applies. Will give zeros, but not every zero. So it's a bad
choice.
c.x|x=k/4 pi for every integer k
* Perfect!, k=0 gives x=0. k=1 gives x=pi, k=2 gives x=2pi, etc. So this is the correct choice in that it gives every possible 0 for each value of k.
d. x|x=6k for every integer k thanks !
* Same problem. This gives zeros, but not every zero.
