All categories

4 years ago

If a pair of perpendicular lines are rotated, which is true?

Mathematics

2 answers:

larisa [96]4 years ago

8 0

Rotated perpendicular lines always remain perpendicular lines.

Rzqust [24]4 years ago

5 0

The <em>correct answer</em> is:

D) Rotated perpendicular lines always remain perpendicular lines.

Explanation:

Perpendicular lines, by definition, meet at a 90° angle.

As long as both lines are rotated the same direction and the same angle, then the angle between the two lines will not change. They will continue to be perpendicular, no matter how many degrees or which direction they are rotated.

You might be interested in

What is the 2nd term when −3f3 + 9f + 6f4 − 2f2 is arranged in descending order?

Brilliant_brown [7]

For this case we have the following polynomial:

$-3f ^ 3 + 9f + 6f ^ 4 - 2f ^ 2$

To answer the question, what we must do is rewrite the polynomial in its standard form.

We have then that the polynomial will be given by:

$6f ^ 4 - 3f ^ 3 - 2f ^ 2 + 9f$

Therefore, we have the ordered polynomial in descending form of exponents.

Therefore, the second term of the polynomial is:

$- 3f ^ 3$

Answer:

The second term of the polynomial is given by:

$- 3f ^ 3$

6 0

3 years ago

3(4x +5y) i need help on this lol thanks

Digiron [165]

Answer: 12x+15y

Step-by-step explanation:

5 0

3 years ago

The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%

liq [111]

<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by $\frac{100+1.8}{100}$ = $\frac{101.8}{100}$ .

So, the population in the year t can be given by $P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}$

Population in the year 2000 = $3,381,000\textrm{x}(\frac{101.8}{100})^{6}$ = $3,762,979.38$

Population in year 2000 = 3,762,979

Let us assume population doubles by year $y$ .

$2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}$

$log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})$

$y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537$

$y$ ≈ $2033$

∴ By 2033, the population doubles.

4 0

4 years ago

Task 1: Billy is trying to download a several apps for his iPhone. His iPhone’s storage capacity is 32GB. The apps he wants

Roman55 [17]

Answer:

Billy can download 6 apps

3 0

2 years ago

Read 2 more answers

How do I long divide ?

katovenus [111]

If you are dividing numbers than here are some steps.

<span>Divide.
Multiply & subtract.
Drop down the next digit. h t o. ) 2 7 8.
Two goes into 2 one time, or 2 hundreds ÷ 2 = 1 hundred. h t o. ) 2 7 8. - ...Divide.Multiply & subtract.<span>Drop down the next digit. h t o. 1 3 9. ) 2 7 8. - 0 7. - 1 8.

Hope this helps! I used this to teach myself.</span></span>

3 0

3 years ago