All categories

3 years ago

Why is the information in the diagram enough to determine that △LMN ~ △PON using a rotation about point N and a dilation?

Mathematics

1 answer:

nekit [7.7K]3 years ago

3 0

Answer:

Step-by-step explanation:

Ed2020

You might be interested in

Phil received a prize of xxx dollars from a poker tournament. The tournament cost him 100100100 dollars to enter.

nikitadnepr [17]

Answer:

$(x-100)

Step-by-step explanation:

Phil entered the poker tournament with $100.
He received a price of $x.

His net winnings from the tournament will be the price minus the entry fee.

Therefore:

Phil's net winnings=$(x-100)

8 0

4 years ago

If line segment AB measures approximately 8.6 units and is considered the base of parallelogram ABCD, what is the approximate co

liraira [26]

Answer:

about 4.8 units

Step-by-step explanation:

6 0

2 years ago

I am in algebra 1, please help i am so lost :(

SSSSS [86.1K]

Doug is 5 years old.
5 + 3 = 8
5 × 8 = 40

3 0

3 years ago

Plz help me with this

AnnZ [28]

Answer: $\bold{A)\quad y=5sin\bigg(\dfrac{6}{5}x-\pi\bigg)-4}$

<u>Step-by-step explanation:</u>

The general form of a sin/cos function is: y = A sin/cos (Bx-C) + D where the period (P) = 2π ÷ B

In the given function, $B=\dfrac{6}{5}$ → $P=2\pi \cdot \dfrac{5}{6}=\dfrac{10\pi}{3}$

Half of that period is: $\dfrac{1}{2}\cdot \dfrac{10\pi}{3}=\large\boxed{\dfrac{5\pi}{3}}$

Calculate the period for each of the options to find a match:

$A)\quad B=\dfrac{6}{5}:\quad 2\pi \div \dfrac{6}{5}=2\pi \cdot \dfrac{5}{6}=\dfrac{5\pi}{3}\quad \leftarrow\text{THIS WORKS!}\\\\\\B)\quad B=\dfrac{6}{10}:\quad 2\pi \div \dfrac{6}{10}=2\pi \cdot \dfrac{10}{6}=\dfrac{10\pi}{3}\\\\\\C)\quad B=\dfrac{5}{6}:\quad 2\pi \div \dfrac{5}{6}=2\pi \cdot \dfrac{6}{5}=\dfrac{12\pi}{5}\\\\\\D)\quad B=\dfrac{3}{10}:\quad 2\pi \div \dfrac{3}{10}=2\pi \cdot \dfrac{10}{3}=\dfrac{20\pi}{3}$

7 0

3 years ago

Solve the inequality 3(2x+7)>-3+5x.

lilavasa [31]

Answer:

Check pdf

Step-by-step explanation:

Download pdf

7 0

3 years ago