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IgorC [24]
3 years ago
12

Use scalar multiplication to determine the coordinates of the vertices of the dilated figure. Then graph the pre-image and the i

mage of the same coordinate grid.

Mathematics
1 answer:
Luba_88 [7]3 years ago
3 0

Answer:

The coordinates of the vertices of the dilated figure are:

A' is (-2 , 4), B' is (4 , 8), C' is (4 , -2), D' is (-2 , -6) ⇒ the answer is (d)

Step-by-step explanation:

* Lets study the matrix of the dilation

- If we dilate any point by scale factor k we  multiply the

 coordinates of the point by k

- The matrix of the dilation by scale factor k is

 \left[\begin{array}{ccc}k&0\\0&k\end{array}\right]

* Now lets solve the problem

- We will multiply the matrix of dilation by the matrix of the

  vertices of the quadrilateral

- The dimension of the matrix of the dilation is 2×2 and the

  dimension of the matrix of the vertices of the quadrilateral

  is 2×4 then the dimension of the matrix of the image of the

  quadrilateral is 2×4

∵ The scale factor is 2

∴ The matrix of dilation is \left[\begin{array}{cc}2&0\\0&2\end{array}\right]

∵ The matrix of the vertices of the quadrilateral is

  \left[\begin{array}{cccc}-1&2&2&-1\\2&4&-1&-3\end{array}\right]

∴ The image of the quadrilateral is :

  \left[\begin{array}{cc}2&0\\0&2\end{array}\right]\left[\begin{array}{cccc}-1&2&2&-1\\2&4&-1&-3\end{array}\right]=

  \left[\begin{array}{cccc}(2)(-1)+(0)(2)&(2)(2)+(0)(4)&(2)(2)+(0)(-1)&(2)(-1)+(0)(-3)\\(0)(-1)+(2)(2)&(0)(2)+(2)(4)&(0)(2)+(2)(-1)&(0)(-1)+(2)(-3)\end{array}\right]=

 \left[\begin{array}{cccc}-2&4&4&-2\\4&8&-2&-6\end{array}\right]

∴ The image of point A' is (-2 , 4)

∴ The image of point B' is (4 , 8)

∴ The image of point C' is (4 , -2)

∴ The image of point D' is (-2 , -6)

* The right answer is figure (d)

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

Here we have , a(t) = (t - k)(t - 3)(t - 6)(t + 3) is a polynomial function of t, where k is a constant.  Given that a(2) = 0 . We need to find the  absolute value of the product of the zeros of a . Let's find out:

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a) dx/dt = 600 - 6x

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c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

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b) dC/dt = 24 - 6C

dC/(24 - 6C) = dt

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So, the equation describing concentration of salt at anytime in the tank is

In (24 - 6C) = K - 6t

In (24 - 6C) = 3.02 - 6t

24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾

6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)

C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)/6

C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

But C = x/25

x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

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20 = 100 - 4.12(e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

- (6t - 3.02) = In 80

- (6t - 3.02) = 4.382

(6t - 3.02) = -4.382

6t = -4.382 + 3.02

t = 1.362/6 = 0.227 min = 13.62 s

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4 years ago
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