Answer:
AC and OA
Step-by-step explanation:
-A secant is a line connecting two points on the circle.
-Given the square OABC of sides 6cm and a circle of r=5cm and the center of the circle as O, and that the radius of the circle is less than the side of the square:
-The circle passes through OA and OC, but doesn't pass through AB and BC.
Hence, AC and OA are the circle's secants.
Answer:
See below
Step-by-step explanation:
<em>Refer to attached. All answers included.</em>
- <u>Note. Will mainly refer to y-coordinates in distance questions</u>
<h3>Part A</h3>
- Distance from coffee shop to library: 16 -8 = 8
- Distance from coffee shop to school: 16 - 10 = 6 = x
- Distance from school to library: 10 - 8 = 2 = y
- Ratio x:y = 6: 2 = 3:1
<h3>Part B</h3>
- Distance from Sergio's home to Saskia's home: 19 -3 = 16
- Distance from Sergio's home to Post office: 16*1/4 = 4
- Coordinates of post office: x = 16+4/4= 17, y = 19-16/4= 15
<h3>Part C</h3>
- Distance from Kay's home to twin theater: 20 - 4 = 16
- Distance from twin theater to market: 16*3/4 = 12
- Coordinates of market: x = 5+4/4= 6, y = 4 + 16/4= 8
Answer:
Jenn simple interest was 1.5% annually making Jenns account have the highest interest rate.
Step-by-step explanation:
Let stadium 1 be the one on the left and stadium 2 the one on the right.
Angle above stadium 1 is 72.9° and the angle above stadium 2 is 34.1° using the angle property of alternate angles(because both the ground and the dotted line are parallel).
For the next part we need to use the trigonometric function of tangent.
As tan x = opposite / adjacent,
Tan 72.9°=1500/ adjacent ( the ground from O to stadium 1)
Therefore the adjacent is 1500/tan 72.9°= 461.46 m( to 5 s.f.)
Same for the next angle,
Tan 34.1°=1500/ adjacent ( the ground from O to stadium 2)
Therefore, the adjacent is 1500/tan 34.1° = 2215.49 m (to 5 s.f.)
Thus, the distance between both stadiums is 2215.49-461.46= 1754.03 m
Correcting the answer to whole number gives you 1754 m which is the option C.
Answer:
1. 15
2. 8
Step-by-step explanation:
The two sequence are geometric progression GP, because they follow a constant multiple (common ratio)
The nth term of a GP is;
Tn = ar^(n-1)
Where;
a = first term
r = common ratio
For the first sequence;
The common ratio r is
r = T3/T2 = 540/90 = 6
r = 6
T2 = ar^(2-1) = ar
T2 = 90 = ar
Substituting the values of r;
90 = a × 6
a = 90/6
a = 15
First term = 15
2. The sam method applies here.
Common ratio r = T3/T2 = 128/32 = 4
r = 4
T2 = ar^(2-1) = ar
T2 = 32 = ar
Substituting the values of r;
32 = a × 4
a = 32/4
a = 8
First term = 8
