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3 years ago

7

If y varies inversely with x and y= 17 when z = 11 when y = 11 find y when x=19 .

1 answer:

BaLLatris [955]3 years ago

8 0

Where’s your z? incomplete question

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For what values of y:is the value of the fraction 7−2y /6 greater than the value of the fraction 3y−7 /12 ?

Fynjy0 [20]

Let's start by plugging in 1 for y. You get 7-2/6, which is 5/6. The other equation is 3-7/12, which equals -1/3. Now let's try 10. You get 7-2(10)/6, which is -13/6. The other equation is 3(10)-7/12, which is 23/12, and is larger!

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3 years ago

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2 years ago

Maria's fish tank has 17 liters of water in it. She plans to add 6 liters per minute until the tank has more than 47 liters. Wha

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t < 5

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3 years ago

The equation for decibels and sound intensity is D=10log(sub 10) I/I(sub0) where D is decibels.

velikii [3]

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3 years ago

PLEASE HELP/ANSWER! From 1980 to 1990, Lior’s weight increased by 25%. If his weight was k kilograms in 1990, what was it in 198

Elena-2011 [213]

The weight in 1980 is $\frac{4k}{5}$ kilograms

<em><u>Solution:</u></em>

From 1980 to 1990, Lior’s weight increased by 25%

His weight is "k" kilograms in 1990

<em><u>To find: weight in 1980</u></em>

This is a percentage increase problem

Let "x" be the weight in kilograms in 1980

<em><u>The percentage increase is given by formula:</u></em>

$\text{Percentage increase } = \frac{\text{Final value - initial value}}{\text{initial value}} \times 100$

Here,

Initial value in 1980 = x

Final value in 1990 = k

Percentage increase = 25 %

<em><u>Substituting the values in formula,</u></em>

$25 = \frac{k-x}{x} \times 100\\\\25x = 100(k-x)\\\\x = 4(k-x)\\\\x = 4k - 4x\\\\5x = 4k\\\\x = \frac{4k}{5}$

Thus the weight in kilograms in 1980 is $\frac{4k}{5}$

7 0

3 years ago