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3 years ago

11

What is the image of(8,12) after a dilation by a scale factor of 1/4 centered at the origin?

1 answer:

max2010maxim [7]3 years ago

3 0

<h3>Answer: (2, 3)</h3>

=================================================

Explanation:

1/4 = 0.25 is the scale factor

Multiply this with each coordinate of the given point

0.25*8 = 2 is the new x coordinate

0.25*12 = 3 is the new y coordinate

So (8,12) moves to (2,3) after applying the dilation

The scale factor k makes 0 < k < 1 true, so the point is closer to the origin after applying the dilation.

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Side note: this trick of multiplying the scale factor by each coordinate only works if the dilation is centered at the origin. For any other center, you'll need to apply a translation first, dilate, then translate back again.

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Solve x2 + 1/2x +1/16=4/9

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<u>Answer:</u>

x = 0.417 or x = -0.917

<u>Step-by-step explanation:</u>

We are given the following expression and we are to solve it for the variable x:

$x ^2 + \frac { 1 } { 2 } x + \frac { 1 } { 1 6 } = \frac { 4 } { 9 }$

We will find the least common multiple of 2. 6 and 9:

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$144x^2+72x+9-64=0$

$144x^2+72x-55=0$

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$x=\frac{-b \pm\sqrt{b^2-4ac} }{2a}$

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3 years ago

Find three positive numbers whose sum is 140 and whose product is a maximum. (Enter your answers as a comma-separated list.)

Paladinen [302]

Answer:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

Step-by-step explanation:

Let the numbers be $x, y\ and\ z.$

Such that:

$x + y + z = 140$

Make z the subject

$z = 140 -x - y$

For their product to be maximum, we have:

$f(x,y,z) = xyz$

Substitute $z = 140 -x - y$ in $f(x,y,z) = xyz$

$f(x,y) = xy(140 - x - y)$

Open bracket

$f(x,y) = 140xy - x^2y - xy^2$

Differentiate w.r.t x and y

$f_x=140y - 2xy - y^2$

$f_y=140x - x^2 - 2xy$

Since the products are maximum, then $f_x = f_y = 0$

For $f_x=140y - 2xy - y^2$

$140y - 2xy - y^2 = 0$

Factorize:

$y(140 - 2x - y) = 0$

Split

$y = 0\ or\ 140 - 2x - y = 0$

Make y the subject

$y = 0\ or\ y = 140 - 2x$

For $f_y=140x - x^2 - 2xy$

$140x - x^2 - 2xy = 0$

---------------------------------------------------

Substitute y = 0

$140x - x^2 -2x*0 = 0$

$140x - x^2 = 0$

Factorize

$x(140 - x)= 0$

$x = 0\ or\ 140-x = 0$

$x = 0\ or\ x = 140$

---------------------------------------------------

Substitute $y = 140 - 2x$

$140x - x^2 - 2xy = 0$

$140x - x^2 - 2x(140 - 2x) = 0$

$140x - x^2 - 280x + 4x^2 = 0$

Re-arrange

$4x^2 -x^2 +140x - 280x = 0$

$3x^2 -140x = 0$

Factor x out

$x(3x - 140) = 0$

Divide through by x

$3x - 140 = 0$

$3x = 140$

$x = \frac{140}{3}$

Recall that: $y = 140 - 2x$

$y = 140 - 2 * \frac{140}{3}$

$y = 140 - \frac{280}{3}$

Take LCM

$y = \frac{140*3-280}{3}$

$y = \frac{140}{3}$

Recall that:

$z = 140 -x - y$

$z = 140 - \frac{140}{3} - \frac{140}{3}$

Take LCM

$z = \frac{3 * 140- 140 - 140}{3}$

$z = \frac{140}{3}$

Hence, the numbers are:

$\frac{140}{3}, \frac{140}{3}, \frac{140}{3}$

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