They went on the slide 6 times and they went on the ferries wheel 3 times(1x6=6 2x3=6)
Answer:
<h3><u>Part (a)</u></h3>
<u />
<u>Equation of a circle</u>

where:
- (a, b) is the center
- r is the radius
Given equation: 
Comparing the given equation with the general equation of a circle, the given equation is a <u>circle</u> with:
- center = (0, 0)
- radius =

To draw the circle, place the point of a compass on the origin. Make the width of the compass 2.5 units, then draw a circle about the origin.
<h3><u>Part (b)</u></h3>
Given equation: 
Rearrange the given equation to make y the subject: 
Find two points on the line:


Plot the found points and draw a straight line through them.
The <u>points of intersection</u> of the circle and the straight line are the solutions to the equation.
To solve this algebraically, substitute
into the equation of the circle to create a quadratic:



Now use the quadratic formula to solve for x:



To find the coordinates of the points of intersection, substitute the found values of x into 


Therefore, the two points of intersection are:

Or as decimals to 2 d.p.:
(2.35, -0.85) and (-0.85, 2.35)
Yes I agree 6 5/6 yeet big boi
Just subtract the whole numbers, 5.93-5 = 0.93.
Answer:
The fourth pair of statement is true.
9∈A, and 9∈B.
Step-by-step explanation:
Given that,
U={x| x is real number}
A={x| x∈ U and x+2>10}
B={x| x∈ U and 2x>10}
If 5∈ A, Then it will be satisfies x+2>10 , but 5+2<10.
Similarly, If 5∈ B, Then it will be satisfies 2x>10 , but 2.5=10.
So, 5∉A, and 5∉B.
If 6∈ A, Then it will be satisfies x+2>10 , but 6+2<10.
Similarly, If 6∈ B, Then it will be satisfies 2x>10 , and 2.6=12>10.
So, 6∉A, and 6∈B.
If 8∈ A, Then it will be satisfies x+2>10 , but 8+2=10.
Similarly, If 8∈ B, Then it will be satisfies 2x>10. 2.8=16>10.
So, 8∉A, and 8∈B.
If 9∈ A, Then it will be satisfies x+2>10 , but 9+2=11>10.
Similarly, If 9∈ B, Then it will be satisfies 2x>10. 2.9=18>10.
So, 9∈A, and 9∈B.