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lana66690 [7]
3 years ago
8

The variable z is directly proportional to x, and inversely proportional to y. When x is 20 and y is 6, z has the value 66.66666

6666667. What is the value of z when x= 25, and y= 12
Mathematics
1 answer:
sergejj [24]3 years ago
6 0

Answer:

41.667 to the nearest thousandth.

Step-by-step explanation:

z = kx/y where k is the constant of proportionality.

Inserting the given values:

66.666666666667 = k*20/6

20k = 6*66.666666666667

k = 6*66.666666666667 / 20

=  20

So the equation of variation is z = 20x /y

When x = 25 and y = 12:

z = 20*25/12

= 41.666666666667.

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Which equation represents the circle described?
faust18 [17]
ANSWER

{(x - 4)}^{2} +  {(y - 3)}^{2}  =   {2}^{2}

EXPLANATION

The equation of the circle with radius r and centre (a,b) is given by


(x - a)^{2}  +  {(x - b)}^{2}  =  {r}^{2}

The radius is

r = 2

We need to determine the center of the circle from the given equation of another circle, which is,


{x}^{2}  +  {y}^{2}  - 8x - 6y + 24 = 0


We complete the square to obtain,


{x}^{2}  - 8x+  {y}^{2}  - 6y + 24 = 0

{x}^{2}  - 8x+  {y}^{2}  - 6y  =  - 24
{x}^{2}  - 8x+  {( - 4)}^{2} +   {y}^{2}  - 6y  +  {( - 3)}^{2}  =  - 24 +  {( - 3)}^{2} +  {( - 4)}^{2}


{(x - 4)}^{2} +  {(y - 3)}^{2}  =  - 24 +  9+  16




{(x - 4)}^{2} +  {(y - 3)}^{2}  =  1


The centre of this circle is (4,3)


Hence the center of the circle whose equation we want to find is also (4,3).


With this center and radius 2, the required equation is,


{(x - 4)}^{2} +  {(y - 3)}^{2}  =   {2}^{2}


Therefore the correct answer is C.

3 0
3 years ago
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1.<br> What is the solution?<br> y<br> 3<br> -3-2-1<br> 2 3x<br> O (1,2)<br> O (1,-1)<br> O 1<br> 02
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Answer:

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Step-by-step explanation:

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3 years ago
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nirvana33 [79]

Given:

The function is:

f(x)=-2x+5

To find:

The inverse of the given function, then draw the graphs of function and its inverse.

Solution:

We have,

f(x)=-2x+5

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Step 4: Substitute y=g(x).

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Hence, the inverse of the given function is g(x)=\dfrac{5-x}{2} and the graphs of these functions are shown below.

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