Ask question

Ask question

All categories

3 years ago

12

Explain how you know that each pair of triangles are similar.

1 answer:

xenn [34]3 years ago

8 0

<em>If two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar. We know this because if two angle pairs are the same, then the third pair must also be equal. When the three angle pairs are all equal, the three pairs of sides must also be in proportion.</em>

You might be interested in

Suppose that the Band Booster Club sells 25 blanket wraps during a chilly football game. Use the graph to determine how many T-s

igor_vitrenko [27]

Answer:

Can I see the graph?

Step-by-step explanation:

You need the info of the graph to get the answer.

6 0

2 years ago

X3(2x2 + 3x) <br><br> a. 2x5 + 3x4 <br> b. 5x9 <br> c. 2x6 + 3x3 <br> d. 5x

Serhud [2]

Answer:

$2x^5+3x^4$

Step-by-step explanation:

$x^3\left(2x^2+3x\right)$

<u>Apply distributive law: </u> $a\left(b+c\right)=ab+ac$

$x^3\left(2x^2+3x\right)=x^3\times \:2x^2+x^3\times \:3x$

$x^3\times \:2x^2+x^3\times \:3x$

$2x^5+3x^4$

<u>OAmalOHopeO</u>

5 0

2 years ago

Express the square root of -80 in its simplest terms.

spin [16.1K]

Answer:

C

Step-by-step explanation:

noting that $\sqrt{-1}$ = i

Given

$\sqrt{-80}$

= $\sqrt{16(-1)(5)}$

= $\sqrt{16}$ × $\sqrt{-1}$ × $\sqrt{5}$

= 4i $\sqrt{5}$ → C

4 0

2 years ago

Please please please help me I need it within 10 minutes!

Alja [10]

Mr Green- Brown tie
Mr Black- Green tie
Mr. Brown- Black tie

I'm working on 6

Hope this helps in the meantime :)

5 0

3 years ago

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.

WARRIOR [948]

Answer:

For First Solution: $y_1(t)=e^t$

$y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution: $y_2(t)=cosht$

$y_2(t)=cosht$ is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: $y_1(t)=e^t$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_1(t)=e^t$

First order derivative:

$y'_1(t)=e^t$

2nd order Derivative:

$y''_1(t)=e^t$

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence $y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution:

$y_2(t)=cosht$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_2(t)=cosht$

First order derivative:

$y'_2(t)=sinht$

2nd order Derivative:

$y''_2(t)=cosht$

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence $y_2(t)=cosht$ is the solution of equation y''-y=0.

3 0

2 years ago