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2 years ago

How do I find dis? 180 rotation about the point (1,4)

Mathematics

1 answer:

prohojiy [21]2 years ago

8 0

Answer:

When we rotate a point A(1, 4) 180 degrees counterclockwise about the origin, the coordinates of point A(1, 4) transformed to A'(-1, -4).

Step-by-step explanation:

We know that 180 Degree Rotation.

We know that when we rotate a point, let say P(x, y), 180 degrees counterclockwise about the origin, the coordinates of point P(x, y) transformed to P'(-x, -y).

In other words, the sign of both x and y coordinates are reversed.

Thus, the rule is:

P(x, y) → P'(-x, -y)

Given the point (1, 4)

P(x, y) → P'(-x, -y)

A(1, 4) → A'(-1, -4)

Thus, when we rotate a point A(1, 4) 180 degrees counterclockwise about the origin, the coordinates of point A(1, 4) transformed to A'(-1, -4).

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Answer:

4.7

Step-by-step explanation:

Just do 28/6 and the simplify it to 14/3 and divide and get 4.6 repeating round it to 4.7

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3 years ago

The table below represents an exponential function what is the common ratio

Greeley [361]

Solution:

Given:

A table representing an exponential function.

The x-values represent the terms, while the y-values represent the numbers.

The common ratio is gotten from the numbers (y-values).

The formula for common ratio is given by;

$\begin{gathered} r=\frac{present\text{ term}}{preceed\text{ ing term}} \\ \\ \text{Hence,} \\ r=\frac{108}{36}=\frac{36}{12}=\frac{12}{4}=\frac{4}{1.33\ldots}=\frac{1.33\ldots.}{0.44\ldots} \\ r=3 \end{gathered}$

Therefore, the common ratio is 3.

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1 year ago

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Step-by-step explanation:

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2 years ago

Read 2 more answers

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$