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Darina [25.2K]
3 years ago
12

Two quantities vary inversely. If the value of the first is 15 when the value of the second is 18, find the value of the second

quantity when the first is 10.. .
Mathematics
2 answers:
vovikov84 [41]3 years ago
7 0

Answer:

The value of second quantity is 27

Step-by-step explanation:

Inverse variation can be represented by the equation xy=k or y = \frac{k}{x} i.e,

y varies inversely as x , if there is some nonzero constant k such that, xy =k or y = \frac{k}{x} where  x≠0 and y≠0.

Given: Two quantities vary inversely.

Let x be the first quantity and y be the second quantity if they  vary inversely.

then, by definition of inverse variation;

y = \frac{k}{x}            .....[1]

Substitute the value of x=15 and y=18 to solve for k;

18= \frac{k}{15}

Multiply both sides by 15 we have;

15\times 18=15 \times \frac{k}{15}

Simplify:

k=270

now, find  the second quantity y using same method when x=10.

then;

after substituting the value of x=10 and k=270 in [1] ;

y = \frac{270}{10}

Simplify:

y= 27

Therefore, the second quantity value i.e,y = 27



defon3 years ago
7 0
Y = k/x

15 = k/18

270 = k


10 = 270 /x

10 x =  270

x = 27

the second number is 27

hope this helps
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The profit function for the first version of the device was very similar to the profit function for the new version. As a matter
NeTakaya

Answer:

a) - Compressing the P(new) function by a scale of 0.5 about the y axis.

- Moving the P(new) function down by 104 units.

b) The two simplified functions for P(original)

-0.08x² + 10.8x – 200.

-0.16x² + 21.6x – 504.

Step-by-step explanation:

Complete Question

An electronics manufacturer recently created a new version of a popular device. It also created this function to represent the profit, P(x), in tens of thousands of dollars, that the company will earn based on manufacturing x thousand devices: P(x) = -0.16x² + 21.6x – 400.

a. The profit function for the first version of the device was very similar to the profit function for the new version. As a matter of fact, the profit function for the first version is a transformation of the profit function for the new version. For the value x = 40, the original profit function is half the size of the new profit function. Write two function transformations in terms of P(x) that could represent the original profit function.

b. Write the two possible functions from part a in simplified form.

Solution

The equation for the new profit function is

P(x) = -0.16x² + 21.6x – 400

At x = 40, the original profit function is half the size of the new profit function

First, we find the value of the new profit function at x = 40

P(x) = -0.16(40)² + 21.6(40) – 400 = 208

Half of 208 = 0.5 × 208 = 104

P(original at x = 40) = P(new at x = 40) ÷ 2

Since we are told that P(original) is a simple transformation of the P(new)

P(original) = P(new)/2 = (-0.16x² + 21.6x – 400)/2 = -0.08x² + 10.8x – 200 ... (eqn 1)

Or, P(original) = 104

-0.16x² + 21.6x – 400 = 104

P(original) = -0.16x² + 21.6x – 400 - 104 = -0.16x² + 21.6x – 504.

So, the two functions that are simple transformations of P(new) to get P(original) are

-0.08x² + 10.8x – 200

Obtained by compressing the P(new) function by a scale of 0.5 about the y axis.

And

-0.16x² + 21.6x – 504.

Obtained by moving the P(new) function down by 104 units.

Hope this Helps!!!

4 0
3 years ago
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