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MAVERICK [17]
2 years ago
14

Beth says that the graph of g(x) = + 1 is a translation of 5 units to the left and 1 unit up of f(x) = . She continues to explai

n that the point (0, 0) on the square root function would be translated to the point (–5, 1) on the graph of g(x). Is Beth’s description of the transformation correct? Explain.
Mathematics
1 answer:
Nata [24]2 years ago
7 0

Beth's description of the transformation is incorrect

<h3>Complete question</h3>

Beth says that the graph of g(x)=x-5+1 is a translation of 5 units to the left and 1 unit up of f(x) = x. She continues to explain that the point (0,0) on the square root function would be translated to the point (-5,1) on the graph of g(x). Is Beth's description of the transformation correct? Explain

<h3>How to determine the true statement?</h3>

The functions are given as:

g(x) = x - 5 + 1

f(x) = x

When the function f(x) is translated 5 units left, we have:

f(x + 5) = x + 5

When the above function is translated 1 unit up, we have:

f(x + 5) + 1 = x + 5 + 1

This means that the actual equation of g(x) should be

g(x) = x + 5 + 1

And not g(x) = x - 5 + 1

By comparison;

g(x) = x - 5 + 1 and g(x) = x + 5 + 1  are not the same

Hence, Beth's description of the transformation is incorrect

Read more about transformation at:

brainly.com/question/17121698

#SPJ1

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Complete question :

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Right triangles abc and dbc with right angle c are given below. If cos(a)=15,ab=12 and cd=2, find the length of bd.
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see the attached figure to better understand the problem

we have that

cos(A)=\frac{1}{5} \\ AB=12\ units\\ CD=2\ units

Step 1

<u>Find the value of AC</u>

we know that

in the right triangle ABC

cos (A)=(AC/AB)\\AC=AB*cos(A)

substitute the values in the formula

AC=12*(1/5)\\ AC=2.4\ units

Step 2

<u>Find the value of BC</u>

we know that

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Applying the Pythagorean Theorem

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Step 3  

<u>Find the value of BD</u>

we know that

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therefore

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<u><em>Answer:</em></u>

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<u><em>Explanation:</em></u>

The diagram representing the question is shown in the attached image

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