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MAVERICK [17]
1 year 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]1 year 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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Which of the following does not belong to the solution set of -3x + 7 &lt; 11?
erik [133]
-3x+7<11

-3x<4

x>-4/3

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With your update:

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2 years ago
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klemol [59]

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3 0
3 years ago
A petrol kiosk p is 12 km due north of another petrol kiosk q. The bearing of a police station r from p is 135 degree and that f
tiny-mole [99]

Answer:

Distance between P and R is 40.15 km.

Step-by-step explanation:

From the picture attached,

Petrol kiosk P is 12 km due North of another petrol kiosk Q.

Bearing of a police station R is 135° from P and 120° from Q.

m∠QPR = 180° - 135° = 45°

m∠PQR = 120°

m∠PRQ = 180° - (m∠QPR +m∠PQR)

             = 180° - (45° + 120°)

             = 180° - 165°

             = 15°

Now we apply sine rule in ΔPQR to measure the distance between P and R.

\frac{\text{sin}(\angle QPR)}{\text{QR}}= \frac{\text{sin}(\angle PQR)}{\text{PR}}=\frac{\text{sin}\angle PRQ}{\text{PQ}}

\frac{\text{sin}(45)}{\text{QR}}= \frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}

\frac{\text{sin}(120)}{\text{PR}}=\frac{\text{sin}(15)}{\text{12}}

PR = \frac{12\text{sin}(120)}{\text{sin}(15)}

PR = 40.15 km

Therefore, distance between P and R is 40.15 km.

8 0
3 years ago
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