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3 years ago

What are the coordinates of each point after quadrilateral ABCD is rotated 270° about the origin?

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

8 0

Answer:

A'(1, -1), B'(3, -2), C'(3, -4), D'(1, -5)

Step-by-step explanation:

The rule for a rotation by 270° about the origin is:

(x, y) → (y, −x)

Coordinates of ABCD:

A(1, 1), B(2, 3), C(4, 3), D(5, 1)

Coordinates of the image:

A'(1, -1), B'(3, -2), C'(3, -4), D'(1, -5)

elena55 [62]3 years ago

7 0

Answer:

A'(1,-1) , B'(3,-2) , C'(3,-4) , D'(1,-5)

These are correct answers hope this helps. :)

Step-by-step explanation:

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3 years ago

Make x the subject of the formula х +12c = 5d 63

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3 years ago

Solve the simultaneous equation 4x-y=9 2x-3y= -23

MrMuchimi

Answer:

x = 5

y = 11

Step-by-step explanation:

y = 4x - 9

2x -3(4x - 9) = -23

2x - 12x + 27 = -23

-10x = -23 -27

-10x = -50

minus cancels out

x = 5

y = 4(5) -9

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3 years ago

A chocolate shop has 32 chocolates with orange filling , 10 chocolates with strawberry filling and 5 chocolates with coconut fil

IceJOKER [234]

Answer:

Step-by-step explanation:

If you add all of the numbers above you will get 47 so

32+10+5=47

and of course there are 32 chocolates with orange filling

so the probability is 32/47

4 0

3 years ago

Suppose that x is a binomial random variable with n=5, p=. 3,and q=. 7.1. Write the binomial formula for this situation and list

Radda [10]

Answer:

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: Any from 0 to 5.

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this question:

$n = 5, p = 0.3, q = 1 - p = 0.7$

$P(X = x) = C_{5,x}.(0.3)^{x}.(0.7)^{5-x}$

Possible values of x: 5 trials, so any value from 0 to 5.

For each value of x calculate p(☓ =x)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.3)^{0}.(0.7)^{5-0} = 0.16807$

$P(X = 1) = C_{5,1}.(0.3)^{1}.(0.7)^{5-1} = 0.36015$

$P(X = 2) = C_{5,2}.(0.3)^{2}.(0.7)^{5-2} = 0.3087$

$P(X = 3) = C_{5,3}.(0.3)^{3}.(0.7)^{5-3} = 0.1323$

$P(X = 4) = C_{5,4}.(0.3)^{4}.(0.7)^{5-4} = 0.02835$

$P(X = 5) = C_{5,5}.(0.3)^{5}.(0.7)^{5-5} = 0.00243$

8 0

4 years ago

What are the coordinates of each point after quadrilateral ABCD is rotated 270° about the origin?​

What are the coordinates of each point after quadrilateral ABCD is rotated 270° about the origin?