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zhuklara [117]
3 years ago
11

( Multiple choice) In the game of miniature golf, the ball bounces off the wall at the same angle it hit the wall. (This is the

angle formed by the path of the ball and the line perpendicular to the wall at the point of contact.) In the diagram, the ball hits the wall at a 40 degree angle. What are the values of x and y?
A) x=50; y=40
B) x=40; y=50
C) x=50; y=50
D) x=40; y=40

Mathematics
2 answers:
attashe74 [19]3 years ago
7 0

Answer:  The correct option is

(C) x=50; y=50.

Step-by-step explanation:  Given that in the game of miniature golf, the ball bounces off the wall at the same angle it hit the wall as shown in the figure.

In the diagram, the ball hits the wall at a 40 degree angle.

We are to find the values of x and y.

From the figure, we note that

x^\circ+40^\circ=90^\circ\\\\\Rightarrow x+40=90\\\\\Rightarrow x=50.

Since the ball bounces off the wall at the same angle it hit the wall, so we must have

90^\circ-y^\circ=40^\circ\\\\\Rightarrow 90-y=40\\\\\Rightarrow y=90-40\\\\\Rightarrow y=50.

Thus, the required values of a and y are 50 and 50.

Option (C) is CORRECT.

Goshia [24]3 years ago
6 0
The answer is A because if you do the 40 dagree and a coodinate plane it forms a triangle.
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Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

Case 1: both balls are white.

At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

\dfrac{w-1}{b+w-1}

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}

Case 2: both balls are black

The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

<h2>(b)</h2>

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of \frac{w}{b+w} of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability \frac{b}{b+w} of picking a black ball with both picks.

This leads to an overall probability of

\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}

Of picking two balls of the same colour.

<h2>(c)</h2>

We want to prove that

\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

Expading all squares and products, this translates to

\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}

As you can see, this inequality comes in the form

\dfrac{x}{y}\geq \dfrac{x-k}{y-k}

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y

And this is our case, because in our case we have

  1. x=b^2+w^2
  2. y=b^2+w^2+2bw so, y has an extra piece and it is larger
  3. k=b+w which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

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Find the actual area of the rectangle shown in the scale drawing below. Use the scale 1 inch = 3 feet
dimulka [17.4K]

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6 0
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The bar chart shows the percent of times each color marble was drawn from a bag after 1000 draws with replacement after each dra
DiKsa [7]

Answer:

Following are the solution to this question:

Step-by-step explanation:

In this question, the graph file is missing so, its solution can be defined as follows:

When the bag includes 20 marbles, each will be drawn 5% of the time on average.  

The bar chart shows that nearly 50% of the workforce red marbles are drawn, and about 25% of the time blue and white marbles are drawn. so, the equation is:

\to \frac{50\%}{5\%}= \frac{50 \times 100}{5\times 100}=\frac{50}{5} = 10 \ red \ marbles\\\\\to \frac{25\%}{5\%}= \frac{25 \times 100}{5\times 100}=\frac{25}{5} = 5\  blue \ marbles

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6 0
3 years ago
the sides of one right triangle are 6,8 and 10.the sides of another right triangle are 10,24 and 26.determine if the triangle ar
valkas [14]
So the ratio must be the same for the triangle to be the same
smallest side of 1 right triangle to smalles side of other right trigle shoudl be same as largest  side of 1 right trignale to largest side of other right triangle and also middle side


6:10 should equal 10:26 should equal 8:24
treat as fraction
6/10=10/26=8/24?
3/5=5/13=1/3
we know that 3/5 is not equal to 1/3 so they are NOT SIMILAR

ratios are
3:5
1:3
5:13


the perimiter of the smaller is 6+8+10=24
perimiter of larger is 10+24+26=60
ratio of perimiters larger to smaller is 60:24=5:2

area is 1/2 base times height
longest side=hypotonuse
therefor other 2 sides must be base and height

area of smaller=8 times 6 times 1/2=24

area lf larger=10 times 24 times 1/2=120
ratio of larger ot smaller is 120:24=5:1



not similar


ratio of perimiter larger to smaler=5:2

ratio oif area larger to smaller=5:1
5 0
4 years ago
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