A number is chosen at random from 2 to 16 inclusive. What is the probability that the number is prime?
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Hi there!
We can find the values of x for which f(x) is decreasing by finding the derivative of f(x):
Taking the derivative gets:
Find the values for which f'(x) < 0 (less than 0, so f(x) decreasing):
0 = -8/x³ - 2
2 = -8/x³
2x³ = -8
x³ = -4
Another critical point is also where the graph has an asymptote (undefined), so at x = 0.
Plug in points into the equation for f'(x) on both sides of each x value to find the intervals for which the graph is less than 0:
f'(1) = -8/1 - 2 = -10 < 0
f'(-1) = -8/(-1) - 2 = 6 > 0
f'(-2) = -8/-8 - 2 = -1 < 0
Thus, the values of x are:
Answer: y= 2x +4
Step-by-step explanation:
1. To be able to write the equation of the line, you want to be able to find the slope first. To do so, rearrange the given equation x+2y=2 into slope-intercept form, which is y=mx+b
First subtract x from both side, which will give us 2y=2-x. Rearrange this to get 2y= -x+2. Then, divide both sides by 2. This will give us y= -1/2x+1
2. Now that you have the equation, look for the slope in the new equation; this will be the m value. In this case, the slope is -1/2. Since we are looking for a line that is perpendicular, we have to change the slope so that it is the opposite reciprocal. The opposite reciprocal of -1/2 is 2, so the slope of the equation we want to find is 2.
3. Next, all we have to do is plug the given ordered pair (-5, -6) and the slope that we found (m=2) into the point-slope equation, which is
That will give us:
y+6 = 2(x+5)
4. Now, solve this equation.
y+6 = 2(x+5) --> distribute the 2 inside the parentheses
y+6 = 2x + 10 --> subtract 6 from both sides
y= 2x +4
Proof of the Law of Sines
The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)
a
sin A
=
b
sin B
=
c
sin C
For more see Law of Sines.
Acute triangles
Draw the altitude h from the vertex A of the triangle
From the definition of the sine function
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Repeat the above, this time with the altitude drawn from point B
Using a similar method it can be shown that in this case
c
sin C
=
a
sin A
Combining (4) and (5) :
a
sin A
=
b
sin B
=
c
sin C
- Q.E.D
Obtuse Triangles
The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.
First the interior altitude. This is the same as the proof for acute triangles above.
Draw the altitude h from the vertex A of the triangle
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Draw the second altitude h from B. This requires extending the side b:
The angles BAC and BAK are supplementary, so the sine of both are the same.
(see Supplementary angles trig identities)
Angle A is BAC, so
sin A =
h
c
or
h = c sin A
In the larger triangle CBK
sin C =
h
a
or
h = a sin C
From (6) and (7) since they are both equal to h
c sin A = a sin C
Dividing through by sinA then sinC:
a
sin A
=
c
sin C
Combining (4) and (9):
a
sin A
=
b
sin B
=
c
sin C
