Ask question

Ask question

All categories

3 years ago

5

HELP I'M TIMED! Reflect triangle ABC over the y-axis. What is the relationship between the segments created if you connect the p

oints to the line of reflection. I WILL GIVE YOU BRAINLIEST!

2 answers:

mylen [45]3 years ago

5 0

Answer:

they are perpendicular to each other

Step-by-step explanation:

4vir4ik [10]3 years ago

3 0

This is your answer thanks!

You might be interested in

Alica wrote these numbers:2 9 15 21 which of alicas numbers is not a composite number

natka813 [3]

Answer:

2, because 2 can only be divided by 1 and itself, and thats what a prime number is

3 0

3 years ago

Read 2 more answers

(2a^3 b^4) ^4= <br> What is the answer to this problem ?

Jet001 [13]

Answer: 16a^12 b^16

Step-by-step explanation:

Everything in the parenthesis to the 4th power.

2a^4= 16a

3*4=12

b^4*4=b^16

3 0

3 years ago

Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

5 0

3 years ago

Describe the graph g < 0.6

padilas [110]

Just asking, what book is it?

5 0

3 years ago

Please helppppppppppppppppp

Anika [276]

The answer will be 40. As it doubles twice during 220mins- 10 to 20 and then 20 to 40.

8 0

3 years ago