Answer:
f(2) = 3
f(5/2) = 5
f(½) = -3
Step-by-step explanation:
Given the linear function f(x) = 4x - 5:
<h3>f(x) = 3</h3>
3 = 4x - 5
3 + 5 = 4x - 5 + 5
8 = 4x
8/4 = 4x/4
2 = x
Therefore, when f(x) = 3, x = 2.
<h3>f(x) = 5</h3>
5 = 4x - 5
5 + 5 = 4x - 5 + 5
10 = 4x
10/4 = 4x/4
5/2 = x
Therefore, when f(x) = 5, x = 5/2.
<h3>f(x) = -3</h3>
-3 = 4x - 5
-3 + 5 = 4x - 5 + 5
2 = 4x
2/4 = 4x/4
½ = x
Therefore, when f(x) = -3, x = ½.
Answer:
I don't give answers sorry.
Step-by-step explanation:
What you would do is take 1 and subract the two probabilites from it. The equation should be
. You can find a common denominator by making it both 15.
Answer:
f(N) = 2+ N/2 + 6N² units of time.
Step-by-step explanation:
Assigning 0 to the variable sum takes one unit of time.
Each time you increment sum by B, you need to call the value of sum, sum it to B and assign it to sum, which takes three units of time in total. You are repeating this process for each value of B which ranges from 1 to 4n and for each value of A which ranges from 1 to n/2. Opening the FOR takes also another unit of time, so, as a result, we have
f(N) 1 + 1 (open the FOR in A)+ N/2*(1 (open the FOR in B) + 4N*3) = 2+ N/2 + 6N² units of time. It has order complexity O(N²).
Answer:
20
Step-by-step explanation:
2, 5, 9, 14, ?, 27.
2,5 the number added is 3
5,9 the number added is 4
9,14 the number added is 5
14,20 the number added is 6
20 , 27 the number added is 7
Answer:
1. Construct a segment XY on the a sketch.
2. Elsewhere on the sketch, draw a line and create a point on the line . Label the point P.
3. Select point P and line segment XY. Construct circle by center and radius.
4. Select the circle and line, then construct point of intersection. Label one of the points of intersection point Q.
5. Select point P and point Q. Construct segment PQ.
6. Select the circle, line, and other point of intersection. Hide objects. Only segment XY and segment PQ should be visible on the sketch.
7. Try dragging point X. Segment PQ should lengthen and shorten as segment XY does.