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jeyben [28]
3 years ago
5

Given the algebraic representation of a linear transformation (x,y) (3x-4, 1/5y-2),what trasformations are occuring

Mathematics
1 answer:
vitfil [10]3 years ago
6 0

Answer:

We can define the algebraic transformations used here as:

Horizontal translation.

For a general point (x, y), a horizontal translation of N units is written as:

(x + N, y)

if N < 0, the translation is to the left

if N > 0, the translation is to the right.

Vertical translation:

For a general point (x, y), a vertical translation of N units is written as:

(x , y + N)

If N < 0, the translation is downwards

if N > 0, the translation is upwards.

Vertical dilation:

For a point (x, y), a vertical dilation of scale factor k is given by:

(y, k*y)

Horizontal dilation:

A dilation of scale factor k is written as:

(x/k, y)

Here we go from:

(x, y) to (3x - 4, (1/5)*y - 2)

So we have:

An horizontal dilation of scale factor 1/3 followed of a horizontal translation of 4 units.

And a vertical dilation of scale factor (1/5) followed by a translation downwards of 2 units.

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Three bags of chips and sodas cost $12.50. Four bags of chips and two sodas cost $10.00. What is the cost of each item? ASAP pls
Helen [10]

Step-by-step explanation:

3c + 3s = 12.50

4c + 2s = 10.00

3S = 12.50-3c

s = 4.17 - c

4c + 2 (4.17-c) = 10.00

2c + 8.34 = 10.00

2c = 1.66

chips = 0.83

soda = 3.34

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A cheese farmer distributes 672 ounces of
TiliK225 [7]

Answer:

there are 42 containers

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3 years ago
Find the expansion of tan x about the point X = 0.
yan [13]

Answer:

f(x) = x +\frac{1}{3}x^{3}+....

Step-by-step explanation:

As per the question,

let us consider f(x) = tan(x).

We know that <u>The Maclaurin series is given by:</u>

f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......

So, differentiate the given function 3 times in order to find f'(x), f''(x) and f'''(x).

Therefore,

f'(x) = sec²x

f''(x) = 2 × sec(x) × sec(x)tan(x)

      = 2 × sec²(x) × tan(x)

f'''(x) = 2 × 2 sec²(x) tan(x) tan(x) + 2 sec²(x) × sec²(x)

       = 4sec²(x) tan²(x) + 2sec⁴(x)

       = 6 sec⁴x - 4 sec² x

We then substitute x with 0, and find the values

f(0) = tan 0 = 0

f'(0) =  sec²0 = 1

f''(0) = 2 × sec²(0) × tan(0) = 0

f'''(0) = 6 sec⁴0- 4 sec² 0 = 2

By putting all the values in the Maclaurin series, we get

f(x) = f(0) + \frac{f^{'}(0)}{1!}\cdot x+ \frac{f^{''}(0)}{2!}\cdot x^{2}+\frac{f^{'''}(0)}{3!}\cdot x^{3}+......

f(x) = 0 + \frac{1}{1}\cdot x+ \frac{0}{2}\cdot x^{2}+\frac{2}{6}\cdot x^{3}+......

f(x) = x +\frac{1}{3}x^{3}+....

Therefore, the expansion of tan x at x = 0 is

f(x) = x +\frac{1}{3}x^{3}+.....

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4 years ago
Let X and Y be the following sets:
VashaNatasha [74]

Answer:

{10;20}

Step-by-step explanation:

X & Y = {10;20}

6 0
3 years ago
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