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Licemer1 [7]
4 years ago
11

What methods could you use to calculate the x-coordinate of the midpoint of a horizontal segment with the endpoints of (-20, 0)

and (20, 0)?. . A. divide 0 by 20 . B. calculate the average of the x-coordinates . C. divide 20 by 2 . D. divide 2 by 20 . E. take the average of the endpoints .
Mathematics
2 answers:
Ronch [10]4 years ago
7 0
The coordinates (abscissa and ordinates) of the midpoint of the line segment is the average of the abscissas and the ordinates, respectively. Therefore, the x - coordinate of the midpoint is equal to -20 + 20 or 0 by 2 which is zero. The answer is letter E. 
marta [7]4 years ago
7 0

Answer:

Option B is correct .

Step-by-step explanation:

Consider the line segment joining points A\left ( x_1,y_1 \right )\,,\,B\left ( x_2,y_2 \right ) . It's midpoint is given by \left ( \frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \right ) .

Midpoint is basically the centre point of a line segment .

Here, given points are \left ( -20,0 \right )\,,\,\left ( 20,0 \right ) .

Consider the following:

\left ( x_1,y_1 \right )=\left ( -20,0 \right )\\\left ( x_2,y_2 \right )=\left ( 20,0 \right )

To find: x-coordinate of the midpoint of a horizontal segment with the endpoints (-20,0) , (20,0) .

Solution:

x-coordinate = \frac{x_1+x_2}{2}

On putting x_1=-20\,,\,x_2=20 , we get

x - coordinate = \frac{-20+20}{2}=0

So, here we calculate the average of x-coordinates .

Option B is correct .

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3 letters without replacement 4 letters A B C D how many ways can this be done if the order of the choices matters
Veseljchak [2.6K]

Answer:

Since the order of choice matters, we will permute the values.                       a bit more explanation for this:

If the order of choice did NOT matter, ABC and BCA will be counted as one since order of choice does NOT matter

Since order of choice does matter, ABC , BCA and CAB are all different possibilities for the arrangement of the same 3 letters

Since we have 3 slots:

___  ___ ___

Now, for the first slot. You can out either one if the 4 alphabets in the first slot since no slot has been used as of now

So:

_<u>4</u>_ ___ ___

**Keep in mind that the 4 is the possible number of values this slot can have**

Now that one slot has been used, one of the 4 alphabets has been used and since we are not allowed to repeat the same alphabets, we are left with  3 more alphabets

we can put any one of the 3 alphabets in this second slot, Hence:

_<u>4</u>_ <u>_3_</u> ___

Now that 2 of the 4 alphabets have been used, we are left with only 2 alphabets, so there are only 2 possible alphabets for slot 3

Therefore:

_<u>4</u>_ _<u>3</u>_ _<u>2</u>_

Now that we know the possible alphabets for all 3 slots, we will multiply them with each other to get the total possible number of 3 - alphabet words we can make with 4 alphabets

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We could've used the formula for Permutation as well

8 0
3 years ago
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aleksley [76]
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7 0
3 years ago
Help please !!! At a local Brownsville play production, 420 tickets were sold. The ticket prices varied on the seating
Radda [10]

Answer:

n = 12$    x = 8$   y = 10$     where n, x, and y are number of tickets

12 n + 8 x + 10 y = 3920     and n + x + y = 420

12n + 8 (x + y) + 2 y = 3920

12 n + 8 (5 n) + 2 y = 3920        since 5 (x + y) = n

52 n + 2 y = 3920   or  y = 1960 - 26 n

Also, n + x + y = 420   or n + 5 n = 420   since x + y = 5 n

n = 70    so 70 of the $12 were sold

And since y = 1960 - 26 n     we have y = 140 tickets

Now 12 * 70 + 8 x + 140 * 10 = 3920

This gives x = 210 tickets

Check:   210 + 140 + 70 = 420 tickets

Also, 12 * 70 + 210 * 8 + 140 * 10 = 3920

7 0
3 years ago
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AlexFokin [52]

Step-by-step explanation:

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