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hram777 [196]
4 years ago
10

On a coordinate plane, point A is at (3, 3) and point B is at (negative 3, 3). Point B is the image of point A when A is rotated

90° counterclockwise around the origin. Point B must be the same from the origin as point A.

Mathematics
1 answer:
Vladimir [108]4 years ago
7 0

Answer:

Point B must be at the same distance from the the origin as point A.

Step-by-step explanation:

Coordinates of point A = (3, 3)

When point A is rotated 90° counterclockwise around the origin, coordinates of the new point B will be (-3, 3)

Distance from origin to point A = \sqrt{(3)^{2}+(3)^2}=3\sqrt{2}

Similarly, distance of point B from the origin = \sqrt{(3)^{2}+(3)^2}=3\sqrt{2}

Therefore, distances of both the points from the origin are same.

Point B must be at the same distance from the the origin as point A.

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Which expression is equal to 2sqrt-32?
wariber [46]

Answer:

a

Step-by-step explanation:

2\sqrt{-32} \\=2\sqrt{(-1)(32)} \\=2\sqrt{-1}\sqrt{2^5} \\=2\sqrt{-1} \sqrt{(2^4)(2)} \\=2(2^2)i\sqrt{2} \\=8i\sqrt{2}

8 0
3 years ago
Urn 1 contains two white balls and one black ball, while urn 2 contains one whiteball and ve black balls. One ball is drawn at r
matrenka [14]

Answer:

4/5 is the probability of white ball from urn 2

Step-by-step explanation:

Correct statement of the question is ;

Urn 1 contains two white balls and one black ball, while urn 2 contains one white bait and five black balls. One ball is drawn at random from urn 1 and placed in urn 2. A ball is then drawn from urn 2. It happens to be white. What is the probability that the transferred ball was white?

We solve it ;

Probability of taking out balls from urn 1

<u>a) for white ball</u>

P(w_{1} )=\frac{2}{3}

 

<u>b) for balck ball</u>

P(b_{1} )= \frac{1}{3}

If one ball is taken from urn 1 and put in urn 2 (which readily contains one white and five black balls)

Probability of taking out balls from urn 2

<u>a) for white ball</u>

P(w_{2} )=P(\frac{w_{2} }{w_{1} } ).P(w_{1} )+P(\frac{w_{2} }{b_{1} } ).P(b_{1} )

=\frac{2}{7} . \frac{2}{3} + \frac{1}{7}. \frac{1}{3} =\frac{5}{21}

Then the probability that white ball was transferred from urn 1 to urn 2 is;

P(\frac{white ball transfer }{w_{2} } )= \frac{P(\frac{w_{2} }{w_{1} } ).P(w_{2} )}{P(w_{2}) }

=\frac{ \frac{2}{7}.\frac{2}{3}  }{\frac{5}{21} }

=\frac{4}{5}

So, the probability that white ball is drawn from urn 2 is 4/5

4 0
3 years ago
Divide. Show your work.<br><br> (6x^4-5x^3+6x^2)/2x^2
serg [7]
We can do like we did before, distribute the denominator,

\bf \cfrac{6x^4-5x^3+6x^2}{2x^2}\implies \cfrac{6x^4}{2x^2}-\cfrac{5x^3}{2x^2}+\cfrac{6x^2}{2x^2}\implies \cfrac{6}{2}\cdot \cfrac{x^4}{x^2}-\cfrac{5}{2}\cdot \cfrac{x^3}{x^2}+\cfrac{6}{2}\cdot \cfrac{x^2}{x^2}&#10;\\\\\\&#10;3 x^4x^{-2}-\cfrac{5}{2}x^3x^{-2}+3x^2x^{-2}\implies 3x^{4-2}-\cfrac{5}{2}x^{3-2}+3x^{2-2}&#10;\\\\\\&#10;3x^2-\cfrac{5}{2}x+3x^0\implies 3x^2-\cfrac{5}{2}x+3
6 0
3 years ago
I need help please ASAP and I will mark branlist
Lady_Fox [76]
7- B
8- C
I’m pretty sure please don’t hurt me if I’m wrong
8 0
4 years ago
Suppose 7x+14 ice cream cones were sold on Saturday and 6x-8 were sold on sunday what is the total number of ice cream cones sol
IrinaVladis [17]
The answer would be 13x+6.
(7x+14)+(6x-8)
Combine like terms to get 13x+6
4 0
4 years ago
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