
I have attached the graph
Answer:
7
Step-by-step explanation:

This is the sum of the first three terms of a geometric sequence, where the first term is 4 and the common ratio is ½.
We can use a formula to find the sum, or, since there's only three terms, we can find the value of each term then add up the results.
4 · (½)¹⁻¹ = 4
4 · (½)²⁻¹ = 2
4 · (½)³⁻¹ = 1
4 + 2 + 1 = 7
Statement Problem: P, Q and R are three buildings. A car began its journey at P, drove to Q, then to R and returned to P. The bearing of Q from P is 058° and R is due east of Q. PQ = 114 km and QR = 70 km. ( Draw a clearly labelled diagram to represent the above information.
Solution:
10C4 represent the Combinations of 10 objects taken 4 at a time.
The formula for nCr is:

In the given case, n=10 and r=4. Using these values we get:

This means we can make 210 combinations of 10 objects taken 4 at a time.
Answer:
31/6
Step-by-step explanation:
I got it right on my quiz