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Ierofanga [76]
4 years ago
14

Square ABCD has a side length of 1. Point E lies on the interior of ABCD, and is on

Mathematics
1 answer:
vichka [17]4 years ago
4 0

Answer:

  1 -(√2)/2 ≈ 0.292893

Step-by-step explanation:

The diagonal length AC is √2, so the distance EC is √2 -1. That is the diagonal of the square whose side length is the length of interest. That side length will be (√2)/2 times the length of EC.

  shortest distance from E = (√2 -1)(√2)/2 = 1 -(√2)/2

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Help me with those two questions
Ksenya-84 [330]

Given:

A figure in which a transversal line intersect two parallel lines.

m\angle 2=4x+7, m\angle 7=5x-13, m\angle 5=68 and m\angle 3=3y-2.

To find:

The value of x and y.

Solution:

We know that, if a transversal line intersect two parallel lines, then

(1) Alternate exterior angles are equal.

(2) Same sided interior angles are supplementary. So their sum is 180 degrees.

In the given figure j and k are parallel lines and l is a transversal line.

From the given figure, it is clear that,

m\angle 2=m\angle 7                (Alternate exterior angles are equal)    

4x+7=5x-13

7+13=5x-4x

20=x

Therefore, the value of x is 20.

Now,

\angle 3+\angle 5=180     (Same sided interior angles are supplementary)

3y-2+68=180

3y+66=180

3y=180-66

y=\dfrac{180-66}{3}

y=\dfrac{114}{3}

y=38

Therefore, the value of y is 38.

3 0
3 years ago
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mixas84 [53]

Answer:

student a because the line has a negative slope not a positive slope

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
A rat is trapped in a maze. Initially he has to choose one of two directions. If he goes to the right, then he will wander aroun
Brut [27]

Answer:

The expected number of minutes the rat will be trapped in the maze is 21 minutes.

Step-by-step explanation:

The rat has two directions to leave the maze.

The probability of selecting any of the two directions is, \frac{1}{2}.

If the rat selects the right direction, the rat will return to the starting point after 3 minutes.

If the rat selects the left direction then the rat will leave the maze with probability \frac{1}{3} after 2 minutes. And with probability \frac{2}{3} the rat will return to the starting point after 5 minutes of wandering.

Let <em>X</em> = number of minutes the rat will be trapped in the maze.

Compute the expected value of <em>X</em> as follows:

E(X)=[(3+E(X)\times\frac{1}{2} ]+[2\times\frac{1}{6} ]+[(5+E(X)\times\frac{2}{6} ]\\E(X)=\frac{3}{2} +\frac{E(X)}{2}+\frac{1}{3}+\frac{5}{3} +\frac{E(X)}{3} \\E(X)-\frac{E(X)}{2}-\frac{E(X)}{3}=\frac{3}{2} +\frac{1}{3}+\frac{5}{3} \\\frac{6E(X)-3E(X)-2E(X)}{6}=\frac{9+2+10}{6}\\\frac{E(X)}{6}=\frac{21}{6}\\E(X)=21

Thus, the expected number of minutes the rat will be trapped in the maze is 21 minutes.

3 0
4 years ago
Find the rule and the graph of the function whose graph can be obtained by performing the translation 3 units left and 2 units d
ahrayia [7]

Answer: Option b.

Step-by-step explanation:

 1. You have the following parent function given in the problem above:

f(x)=x³ (This is the simplest form. We need to translate it 3 units left and 2 units down)

2. If you take the parent function and make y=f(x+3), then you have:

y=f(x+3)=(x+3)^{3} (The function is shifted 3 units left on the x-axis).

3. Then you if you make y=f(x+3)-2, as following, you obtain:

y=f(x+3)-2=(x+3)^{3}-2  (The function is shifted 2 units down on the y-axis).

4. Therefore, that is how you obtain the final function.

The answer is the graph shown in the option b.

5 0
3 years ago
7/10 + 7/100 = 77/100
tensa zangetsu [6.8K]

Answer:

what I'm am supposed to be answering

5 0
2 years ago
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