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3 years ago

Please help?? Algebra 1 adding and subtracting radicals!!

Mathematics

1 answer:

Tcecarenko [31]3 years ago

5 0

Answer: 9√2 + 3√5

Step-by-step explanation:

6√2 + 3√5 + 3√2 = √2(6+3) + 3√5 = 9√2 + 3√5

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JJ: ANswer question for me(Tasha) <br> WHat is 456789x456

Grace [21]

Answer:

208295784

Step-by-step explanation:

456789 x 456 = 208295784

7 0

3 years ago

Read 2 more answers

Write an expression to match the statement 1/5 of the difference between 18 and 11

dalvyx [7]

1/5 (18-11) I believe this is what you mean?

4 0

3 years ago

You are dealt two cards successively without replacement from a standard deck of 52 playing cards. Find the probability that the

Irina-Kira [14]

Answer:

1/52

Step-by-step explanation:

52/4=13 theirs only 4 5s in a deck of cards so 13*4 is 52

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3 years ago

I need this one please I don’t understand it

Dovator [93]

Answer:

Hi! your answer should be C. √34

Step-by-step explanation:

a^2+b^2=c^2 (leg^2+leg^2= hypotenuse^2)

5^2+3^2=c^2

25+9=c^2

34=c^2

square root both sides.

√34=c

the hypotenuse is C. √34

5 0

3 years ago

100 points+Brainliest for correct answer

goblinko [34]

Answer:

The electron moved down to an energy level and has an energy of

$10.92 \times 10^{-19} \textsf{J}$

Step-by-step explanation:

$\textsf{initial energy level of electron}= 16.32 \times 10^{-19} \textsf{J}$

$\triangle E =\textsf{energy released}= 5.4 \times 10^{-19} \textsf{J}$

A positive energy change means that the electron absorbs energy, while a negative energy change implies a release of energy from the electron.

Therefore, when an electron absorbs energy, the electron moves up to higher energy levels, and when an electron releases energy, it moves to lower energy levels.

$\triangle E=E_2-E_1$

$\implies 5.4 \times 10^{-19} = 16.32 \times 10^{-19} -E_1$

$\implies E_1= 16.32 \times 10^{-19} -5.4 \times 10^{-19}$

$\implies E_1= 10.92 \times 10^{-19}$

<u>Answer</u>:

The electron moved down to an energy level and has an energy of

$10.92 \times 10^{-19} \textsf{J}$

6 0

2 years ago