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2 years ago

How can a sequence of transformations be used to justify whether two figures are congruent or similar?

Mathematics

1 answer:

Lena [83]2 years ago

3 0

Answer:

A sequence of transformation can move two congruent figures to the same place (vertical / horizontal shifts). This causes them to perfectly overlap, proving their congruency.

A sequence of transformations can be used to show that something is similar, because if one were to transform a figure with only shrinks / stretches, one can shrink / stretch them to make them congruent. If only shrinks or stretches were done, and if the result of the transformation is congruent to the other figure, then the figures are similar (sides are proportional and shrinking/stretching them proves proportionality to other figure).

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otez555 [7]

The answer is 9628528037

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2 years ago

Read 2 more answers

Graphing polynomial functions?

Leni [432]

NOTES:

Degree: the largest exponent in the polynomial

End Behavior:

Coefficient is POSITIVE, then right side goes to POSITIVE infinity
Coefficient is NEGATIVE, then right side goes to NEGATIVE infinity
Degree is EVEN, then left side is SAME direction as right side
Degree is ODD, then left side is OPPOSITE direction as right side

Multiplicity (M): the exponent of the zero. e.g. (x - 3)² has a multiplicity of 2

Relative max/min: the y-value of the vertices.

Find the axis of symmetry (the midpoint of two neighboring zeros)
Plug the x-value from 1 (above) into the given equation to find the y-value. (which is the max/min)
Repeat 1 and 2 (above) for each pair of neighboring zeros.

Rate of Change: slope between the two given points.

********************************************************************************************

1. f(x) = (x-1)²(x + 6)

a) Degree = 3

b) end behavior:

Coefficient is positive so right side goes to positive infinity
Degree is odd so left side goes to negative infinity

c) (x - 1)²(x + 6) = 0

x - 1 = 0 x + 6 = 0

x = 1 (M=2) x = -6 (M=1)

d) The midpoint between 1 and -6 is -3.5, so axis of symmetry is at x = -3.5

y = (-3.5 - 1)²(-3.5 + 6)

= (-4.5)²(2.5)

= 50.625

50.625 is the relative max

e) see attachment #1

f) The interval at which the graph increases is: (-∞, -3.5)U(1, ∞)

g) The interval at which the graph decreases is: (-3.5, 1)

h) f(-1) = (-1 - 1)²(-1 + 6)

= (-2)²(5)

= 20

f(0) = (0 - 1)²(0 + 6)

= (-1)²(6)

= 6

Find the slope between (-1, 20) and (0, 6)

m = $\frac{20-6}{-1-0}$

= $\frac{14}{-1}$

= -14

********************************************************************************************

2. y = x³+3x²-10x

= x(x² + 3x - 10)

= x(x + 5)(x - 2)

a) Degree = 3

b) end behavior:

Coefficient is positive so right side goes to positive infinity

Degree is odd so left side goes to negative infinity

c) x(x + 5)(x - 2) = 0

x = 0 x + 5 = 0 x - 2 = 0

x = 0 (M=1) x = -5 (M=1) x = 2 (M=1)

d) The midpoint between -5 and 0 is -2.5, so axis of symmetry is at x = -2.5

y = -2.5(-2.5 + 5)(-2.5 - 2)

= -2.5(2.5)(-4.5)

= 28.125

28.125 is the relative max

The midpoint between 0 and 2 is 1, so axis of symmetry is at x = 1

y = 1(1 + 5)(1 - 2)

= 1(6)(-1)

= -6

-6 is the relative min

e) see attachment #2

f) The interval at which the graph increases is: (-∞, -2.5)U(1, ∞)

g) The interval at which the graph decreases is: (-2.5, 1)

h) f(-1) = -1(-1 + 5)(-1 - 2)

********************************************************************************************

3. y = -x(x + 2)(x - 7)(x - 3)

a) Degree = 4

b) end behavior:

Coefficient is negative so right side goes to negative infinity

Degree is even so left side goes to negative infinity

c) -x(x + 2)(x - 7)(x - 3) = 0

-x = 0 x + 2 = 0 x - 7 = 0 x - 3 = 0

x = 0 (M=1) x = -2 (M=1) x = 7 (M=1) x = 3 (M=1)

d) The midpoint between -2 and 0 is -1, so axis of symmetry is at x = -1

y = -(-1)(-1 + 2)(-1 - 7)(-1 - 3)

= 1(1)(-8)(-4)

= 32

32 is a relative max

The midpoint between 0 and 3 is 1.5, so axis of symmetry is at x = 1.5

y = -(1.5)(1.5 + 2)(1.5 - 7)(1.5 - 3)

= -1.5(3.5)(-5.5)(-1.5)

= -43.3125

-43.3125 is the relative min

The midpoint between 3 and 7 is 5, so axis of symmetry is at x = 5

y = -(5)(5 + 2)(5 - 7)(5 - 3)

= -5(7)(-2)(2)

= 140

140 is the relative max

e) see attachment #3

f) The interval at which the graph increases is: (-∞, -1)U(1.5, 5)

g) The interval at which the graph decreases is: (-1, 1.5)U(5, ∞)

h) f(-1) = -(-1)(-1 + 2)(-1 - 7)(-1 - 3)

= 1(1)(-8)(-4)

= 32

f(0) = -(0)(0 + 2)(0 - 7)(0 - 3)

= 0

Find the slope between (-1, 32) and (0, 0)

m = $\frac{32-0}{-1-0}$

= $\frac{32}{-1}$

= -32

5 0

3 years ago

Marty is taking piano lessons. She is going to practice every afternoon. The 1st week she practiced for 11 minutes each day. The

defon

Answer:

$m=7w+4$

Step-by-step explanation:

Linear Modeling

It consist is setting up a linear relationship between two variables, given some experimental data. Only 2 points are needed to set up the equation of a line, but if more than 2 points are used, then the result should use statistical approaches like linear regression to find the best-fit line.

For the question at hand, Marty practices his piano lessons 11 minutes the week #1. It provides the first point (1,11). He practices 25 minutes per day on the third week. It gives us another point (3,25). This is enough to find the equation of a line. The general formula for a line, having two points (m1,w1) (m2,w2) is

$\displaystyle m-m_1=\frac{m_2-m_1}{w_2-w_1}(w-w_1)$