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disa [49]
3 years ago
10

Triangle a'b'c' is result of dilating abc about point a by a scale factor of 2/3. Determine whether each claim about the propert

ies of abc and a'b'c' is true or false
Mathematics
2 answers:
3241004551 [841]3 years ago
5 0

Answer:

True , because all properties preserved under dilation from triangle abc to a'b'c'.

Step-by-step explanation:

Given : A triangle abc is dilated by a factor of 2/3 , result triangle a'b'c'

To find :After dilation,  properties of triangle abc and a'b'c' is true or false.

Solution:

          Dilation of a triangle are result either decrease or increase the corresponding length of triangle , its depends on by which scale factor triangle is dilated ,and dilation does not affect the angle measures of triangle.here scale factor is 2/3 so corresponding sides of triangle a'b'c' decreases.

So, dilation of triangle are not congruent but similar to the original triangle abc

Therefore , basic properties of triangle preserved under dilation of triangle by a scale factor.

Reil [10]3 years ago
3 0

Answer:

True

Step-by-step explanation:

Because it is the "<u>Correct Answer</u>"

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anyanavicka [17]

The question is missing parts. Here is the complete question.

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)=xe^{-\frac{x^{2}}{32} } , [ -2,8]

Answer: Absolute maximum: f(4) = 2.42;

              Absolute minimum: f(-2) = -1.76;

Step-by-step explanation: Some functions have absolute extrema: maxima and/or minima.

<u>Absolute</u> <u>maximum</u> is a point where the function has its greatest possible value.

<u>Absolute</u> <u>minimum</u> is a point where the function has its least possible value.

The method for finding absolute extrema points is

1) Derivate the function;

2) Find the values of x that makes f'(x) = 0;

3) Using the interval boundary values and the x found above, determine the function value of each of those points;

4) The highest value is maximum, while the lowest value is minimum;

For the function given, absolute maximum and minimum points are:

f(x)=xe^{-\frac{x^{2}}{32} }

Using the product rule, first derivative will be:

f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )

f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} ) = 0

1-\frac{x^{2}}{16}=0

\frac{x^{2}}{16}=1

x^{2}=16

x = ±4

x can't be -4 because it is not in the interval [-2,8].

f(-2)=-2e^{-\frac{(-2)^{2}}{32} }=-1.76

f(4)=4e^{-\frac{4^{2}}{32} }=2.42

f(8)=8e^{-\frac{8^{2}}{32} }=1.08

Analysing each f(x), we noted when x = -2, f(-2) is minimum and when x = 4, f(4) is maximum.

Therefore, absolute maximum is f(4) = 2.42 and

absolute minimum is f(-2) = -1.76

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3 years ago
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