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alekssr [168]
2 years ago
7

Graph transformations GCSE maths. Please helpppp. Thanks

Mathematics
1 answer:
vitfil [10]2 years ago
3 0
<h3>Answer: Rotate 180 degrees around point B</h3>

Explanation:

Point A(-3,1) moves to A ' (1,1). The line through these two points is y = 1.

Point C(-3,3) moves to C ' (1, -1). The line through these two points is y = -x

These two lines intersect at point B(-1,1) which is the invariant point, aka fixed point. It is the center of rotation.

Notice how points A, B and A' are collinear (i.e. they are on the same line). The same can be said about points C, B and C'.

Because of these two sets of collinear points, it means we have a 180 degree rotation going on.

Check out the diagram below.

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anygoal [31]

Answer:

The vertex is (-1, 4)

Not sure what the table looks like, but just plug the x values into the equation to get a complete point

Graph the points on a graph

Step-by-step explanation:

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2 years ago
How many square inches are 9 cubes with a 1 inch edge
ki77a [65]
Add it up then you will get answer

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what is the equation in point slope form of a line that passes through the point (–8,2) and has a slope of 1/2?
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5 0
3 years ago
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
3 years ago
A projectile’s motion is modeled by the function given in the table, where x represents time in seconds and f(x) represents heig
Pie

Answer: 2 seconds

Step-by-step explanation:

got it right

4 0
2 years ago
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