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Blababa [14]
2 years ago
13

Brenden runs six times a week for thirty minutes and lifts three times a week for twenty minutes. Write a mathematical expressio

n for the number of hours Brenden works out in a week.
Mathematics
1 answer:
hodyreva [135]2 years ago
4 0

Answer: \frac{6(#30)+3(20)}{60}

Step-by-step explanation:

6(30)+3(20)

=180+60

= 240/60

=4 hours

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Find how many six-digit numbers can be formed from the digits 2, 3, 4, 5, 6 and 7 (with repetitions), if:
Goshia [24]

Answer:

case 1 = 2592

case 2 =  729

case 1 + case 2 =  2916

(this is not a direct adition, because case 1 and case 2 have some shared elements)

Step-by-step explanation:

Case 1)

6 digits numbers that can be divided by 25.

For the first four positions, we can use any of the 6 given numbers.

For the last two positions, we have that the only numbers that can be divided by 25 are numbers that end in 25, 50, 75 or 100.

The only two that we can create with the numbers given are 25 and 75.

So for the fifth position we have 2 options, 2 or 7,

and for the last position we have only one option, 5.

Then the total number of combinations is:

C = 6*6*6*6*2*1 = 2592

case 2)

The even numbers are 2,4 and 6

the odd numbers are 3, 5 and 7.

For the even positions we can only use odd numbers, we have 3 even positions and 3 odd numbers, so the combinations are:

3*3*3

For the odd positions we can only use even numbers, we have 3 even numbers, so the number of combinations is:

3*3*3

we can put those two togheter and get that the total number of combinations is:

C = 3*3*3*3*3*3 = 3^6 = 729

If we want to calculate the combinations togheter, we need to discard the cases where we use 2 in the fourth position and 5 in the sixt position (because those numbers are already counted in case 1) so we have 2 numbers for the fifth position and 2 numbers for the sixt position

Then the number of combinations is

C = 3*3*3*3*2*2 = 324

Case 1 + case 2 = 324 + 2592 = 2916

4 0
2 years ago
Which step is the same in the construction of parallel lines and the construction of a perpendicular line through a point on the
madam [21]

Answer:

D.Place the compass on a point on the original line.

Complete Question:

A.Create only one arc that intersects the original line.

B.Create two arcs that intersect the original line.

C.Place the compass on a point off the original line.

D.Place the compass on a point on the original line.

Step-by-step explanation:

* Lets revise the steps of constructing parallel lines and a perpendicular line through a point on the line

<em># Contracting parallel lines</em>

- Given: Line AB and point P not on AB

1. Draw a slant line through point P and intersects AB at point C, this line is the transversal of the parallel lines

2.<u> Place the pin of the compass at point C</u> and draw an arc intersects AB at D and CP at E

3. Without changing the distance of the compass put the pin of the compass on point P and draw an arc intersects CP at point Q

4. Use the compass to measure the distance from D to E and place the pin of the compass at point Q and draw an arc intersects the arc from P to CP at point U

5. Join P and U by a line this line is parallel to AB

<em># Constructing a perpendicular line through a point on the line</em>

- Given line AB and point P lies on it

1. <u>Place your compass pin at P </u>and draw an arc of any size below AB that crosses the line twice at points C and D

2. Stretch the compass to a larger distance

3.<u> </u>Place the compass pin on point C<u> </u>and draw a small arc above the line

4. Without changing the distance of the compass place the compass pin on point D and draw another arc intersects the arc from C at point E

5. Using a straightedge, join P and E where PE is perpendicular to AB

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∵ In first construction we put the pin of the compass at point C which  lies in the original line AB

∵ In second construction we put the pin of the compass at point P which lies in the original line AB

∴ The common step is:  <em><u>"Place the compass on a point on the original line"</u></em>

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