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2 years ago

What set of reflections would carry triangle ABC onto itself?

2 answers:

8 0

The correct answer is:
4) y-axis, x-axis, y-axis, x-axis.

Explanation: 
Reflecting a point (x,y) across the x-axis will map it to (x,-y).
Reflecting a point (x,y) across the y-axis will map it to (-x,y).
Reflecting a point (x,y) across the line y=x will map it to (y, x).

We want a series of transformations that will map every point (x,y) back to (x,y). This means that everything that gets done in one transformation must be undone in another. The only one where this happens is #4.

Reflecting across the y-axis first negates the x-coordinate; (x,y) goes to (-x,y).
Reflecting this across the x-axis negates the y-coordinate; (-x,y) goes to (-x,-y).
Reflecting this point back across the y-axis negates the x-coordinate again, returning it to the original: (-x,-y) goes to (x,-y).
Reflecting this point back across the x-axis negates the y-coordinate again, returning it to the original: (x,-y) goes to (x,y).
We are back to our original point.

8_murik_8 [283]2 years ago

3 0

Answer: 4) y-axis, x-axis, y-axis, x-axis

Step-by-step explanation:

Let (x,y) represents the coordinates of the triangle ABC,

1) Under the set of reflection, x-axis, y=x, y-axis, x-axis,

$(x,y)\rightarrow (x,-y)\rightarrow (-x,-(-y))\rightarrow (-(-x),y)\rightarrow (x,-y)$

Hence, (x,-y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections does not carry triangle ABC onto itself.

2) Under the set of reflection, x-axis, y-axis, x-axis

$(x,y)\rightarrow (x,-y)\rightarrow (-x,-y))\rightarrow (-x,y)$

Hence, (-x,y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections does not carry triangle ABC onto itself.

3) Under the set of reflection, y=x, x-axis, x-axis

$(x,y)\rightarrow (-x,-y)\rightarrow (-x,-(-y))\rightarrow (-x,-y)$

Hence, (-x,-y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections does not carry triangle ABC onto itself.

4) Under the set of reflection, y-axis, x-axis, y-axis, x-axis

$(x,y)\rightarrow (-x,y)\rightarrow (-x,-y)\rightarrow (-(-x),-y)\rightarrow (x,-(-y))$

Hence, (x,y) represents the coordinates of resultant transformed triangle.

⇒ This set of reflections carries triangle ABC onto itself.

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2 years ago

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Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

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Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago

Mason owns a trucking company. For every truck that goes out, Mason must pay the driver $13 per hour of driving and also has an

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Answer: The required system of equations are

y = 30x

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Step-by-step explanation:

Let x represent the number of hours

that the driver worked.

Let y represent the number of miles that the truck drove.

For every truck that goes out, Mason must pay the driver $13 per hour of driving and also has an expense of $1 per mile driven for gas and maintenance. If on a particular day, his total expenses for the driver, gas and truck maintenance were $258, it means that

13x + y = 258

On that particular day, the driver drove an average of 30 miles per hour.

Speed = distance/time

It means that

y/x = 30

y = 30x

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