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murzikaleks [220]

4 years ago

13

The sum of n A.M.'s between two numbers is 20. If the last mean is double of the first and one number is three times the other,

then find the number

1 answer:

tester [92]4 years ago

4 0

Answer:

120

Step-by-step explanation: because you add 20 twice and thts 40 and 40 times 3 is 120

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baherus [9]

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What is 708569 to the nearest ten thousands

gregori [183]

70,000
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Which line is perpendicular to y= -3/4x+ 2 if the line passes through the point (3,5)?

iren2701 [21]

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6 0

3 years ago

Find the value of the variable y, where the sum of the fractions 6/(y+1) and y/(y-2) is equal to their product.

Blizzard [7]

Answer:

The answer is

$y = 3$

$y = - 4$

Step-by-step explanation:

We must find a solution where

$\frac{6}{y + 1} + \frac{y}{y - 2} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

$\frac{6}{y + 1} \times \frac{y - 2}{y - 2} + \frac{y}{y - 2} \times \frac{y + 1}{y + 1}$

Which equals

$\frac{6y - 12}{(y -2) (y + 1)} + \frac{ {y}^{2} + y }{(y - 2)(y + 1)}$

Add the fractions

$\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)} = \frac{6}{y + 1} \times \frac{y}{y - 2}$

Simplify the right side by multiplying the fraction

$\frac{6y}{(y + 1)(y + 2)}$

Set both fractions equal to each other

$\frac{6y}{(y + 1)(y - 2)} = \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}$

Since the denomiator are equal, we must set the numerator equal to each other

$6y = {y}^{2} + 7y - 12$

$= {y}^{2} + y - 12$

$(y + 4)(y - 3)$

$y = - 4$

$y = 3$

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3 years ago

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What is the means-to-MAD ratio of the two data sets, expressed as a decimal?

Ksenya-84 [330]

I think the answer is 1.2

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