3 years ago

The coordinates of the vertices of ARST are

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

3 0

Answer:

A right triangle

You might be interested in

Use Kelly’s steps to simplify this expression.

pantera1 [17]

10w^7 is the answer I’m pretty sure

7 0

3 years ago

Read 2 more answers

The circumference of a circle is 28 x inches. What is the length of the radius of this circle?

meriva

Answer: 14

Explanation: The radius is half of the value of the circumference.

4 0

3 years ago

Giving loads of points to anyone who can answer these questions or even just most of them correctly!!!

cestrela7 [59]

Answer:

below :)

Step-by-step explanation:

1) 12

2}3/2

3)224.64

4)2 quarters

5)w=50

6)m=6y+7

7)42/6=7 x=6

8 0

2 years ago

Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C

netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> $|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}$ </span></span>

Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.

$|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5$

$|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6$

$|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4$

$|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \ \sqrt{25}=5$

$|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4$

Answer: B and D

3 0

3 years ago

4. Which of the following represents the solution set to the inequality

amid [387]

Okay thanks so much for

7 0

2 years ago