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Vlad1618 [11]
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?

Mathematics
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.

--------

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
kolezko [41]

<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:

x^2 \times 144 +\frac{1}{2}x \times 144 +\frac{1}{16} \times 144 = \frac{4}{9} \times 144

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144x^2+72x+9=64

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

144x^2+72x-55=0

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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}

8 0
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