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3 years ago

Which pair of polygons is congruent?

Mathematics

1 answer:

erik [133]3 years ago

4 0

Answer:

Step-by-step explanation:

Polygon 3 and 5 are congruent cause they have the same length of side

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Suppose that a cell phone monthly rate plan costs the user 5 cents per minute beyond a fixed monthly fee of $20. This implies th

nikdorinn [45]

Answer:

y = 20 + .5x

Step-by-step explanation:

Suppose that a cell phone monthly rate plan costs the user 5 cents per minute beyond a fixed monthly fee of $20. This implies that the relationship between monthly cost and monthly number of minutes is linear. Write an equation that relates total monthly cost to monthly minutes used.

Fixed monthly fee = $20

Fee per minute = $0.5

Number of minutes = x

Variable monthly cost = 0.5x

Let Total monthly cost = y

Total monthly cost = fixed monthly cost + variable monthly cost

y = 20 + .5x

Equation that relates total monthly cost to monthly minutes used is y = 20 + .5x

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3 years ago

The entertainment committee at Maplecrest Junior High School has narrowed down the theme of the Spring

Roman55 [17]

Answer:

Assign a number to each student, and use a computer program to generate 100 random numbers between 1 and 2000. Ask those students whose numbers are selected.

Step-by-step explanation:

3 0

4 years ago

Find the values of the six trigonometric functions of an angle in standard position if the point with coordinates (6,-5) lies on

Tom [10]

C i guesss because you just said it

4 0

3 years ago

Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r

erastova [34]

Answer:

The probability is $\frac{56!}{64!}$

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, $64 \choose 8 .$

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function $f : A \rightarrow A ,$ with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

$\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}$

7 0

3 years ago

Pic below................................<br> answer ASAP

Free_Kalibri [48]

Answer:

The answer is C.-216

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers