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NikAS [45]
2 years ago
12

Find the distance between the points given.

Mathematics
1 answer:
Maru [420]2 years ago
4 0

Use the distance formula

d = \sqrt {\left( {x_1 - x_2 } \right)^2 + \left( {y_1 - y_2 } \right)^2

Input corresponding numbers

d = \sqrt {\left( {2- 6} \right)^2 + \left( {5-8} \right)^2

Solve

d = \sqrt {\left( {-4} \right)^2 + \left( {-3} \right)^2

\sqrt{16 + 9}

\sqrt{25} = 5\\\\ d=5

So, the distance between the points given is 5

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x=2 2/3

Step-by-step explanation:

The equation also means that x=8^3

8/3=2 2/3

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A ream of paper containing 500 sheets is 5 cm thick. How many sheets of this type of paper would there be in a stack 7.5 cm high
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Sales variances may be computed in a manner similar to cost variances–that is, computing both price and volume variances.
Sholpan [36]

Answer:

The answer is True.

Step-by-step explanation:

Sales variance is computed in same manner as cost variance that is computing both price and volume variance. However interpretation of end result will not be same. For example in material price variance if

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Material price varaince = 500 (5-4) = 500,

This gives us favourable price variance of 500 dollars. However in sales price variance if

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Y = -2, 3y-4 evaluate it please​
Luba_88 [7]
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a tour-bus operator sold $1,517 worth of tickets for a trip to Niagara Falls. Each tickets cost the same amount and could be pai
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We will make use of the concept of prime factors for solving this question. The prime factors of 1517 are:

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So, as can be clearly seen the Niagara Fall tour operator can either sell 37 tickets for $41 each or 41 tickets for $37 each.

Now, we know that we cannot reach $37 using just three bills. Thus, the only option left to us is $41 which can be broken down into three bills of $20,$20 and $1. This satisfies our condition.

Thus the cost of a ticket was $41 and 37 such tickets were sold.

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2 years ago
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