SA=2lw+2lh+2hw ..........
I believe B or C but that’s a straight guess. i would say B just because if you live like outside of town then ur screwed
Answer:
f'(x) = -1/(1 - Cos(x))
Step-by-step explanation:
The quotient rule for derivation is:
For f(x) = h(x)/k(x)

In this case, the function is:
f(x) = Sin(x)/(1 + Cos(x))
Then we have:
h(x) = Sin(x)
h'(x) = Cos(x)
And for the denominator:
k(x) = 1 - Cos(x)
k'(x) = -( -Sin(x)) = Sin(x)
Replacing these in the rule, we get:

Now we can simplify that:

And we know that:
cos^2(x) + sin^2(x) = 1
then:

So... hmmm if you check the first picture below, for 2)
we could use the proportions of those small, medium and large similar triangles like

now.. for 3) will be the second picture below
Answer:
210
Step-by-step explanation:
Here comes the problem from Combination.
We are being asked to find the number of ways out in which 3 students may sit on 7 seats in a row. Please see that in this case the even can not be repeated.
Let us start with the student one. For him all the 7 seats are available to sit. Hence number of ways for him to sit = 7
Let us see the student second. For him there are only 6 seats available to sit as one seat has already been occupied. Hence number of ways for him to sit = 6
Let us see the student third. For him there are only 5 seats available to sit as two seat has already been occupied. Hence number of ways for him to sit = 5
Hence the total number of ways for three students to be seated will be
7 x 6 x 5
=210