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crimeas [40]
3 years ago
5

Steve had 484848 chocolates but decided to give 888 chocolates to each of his fff coworkers.

Mathematics
1 answer:
katrin2010 [14]3 years ago
7 0

Answer:

483960

Step-by-step explanation:

484848

-     888

______

483960

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Write a story problem about 40 apples and 17 peat
MaRussiya [10]

Answer:

Lets say you have 40 apples and 17 "peats". You decide to give your friend, Timmy, 2/3's of the apple, because you know that he loves apples! You also give him 2/7 of your "peats". How many apples and peats, combined, does that leave you with?

Step-by-step explanation:

There really is none.

4 0
3 years ago
John wants to buy a new game for his computer. He can save $12 a week if he mows his neighbor's lawn and $10 a week if he babysi
kykrilka [37]
It depends on how much the computer game costs.
4 0
3 years ago
Please help me to prove this!<br>I need is no.(c). So, please help me do it.<br>​
zloy xaker [14]

Answer:  see proof below

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

Given: A + B + C = 90°                  → A + B = 90° - C

                                                     → C = 90° - (A + B)

Use the Double Angle Identity:      cos 2A = 1 - 2 sin² A

                                                       → sin² A = (1 - cos 2A)/2

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B

Use the Cofunction Identities:      cos (90° - A) = sin A

                                                       sin (90° - A) = cos A

<u>Proof LHS → RHS:</u>

LHS:                       sin² A + sin² B + sin² C

\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C

\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C

Given:                1 - cos (90° - C) · cos (A - B) + sin² C

Cofunction:       1 - sin C · cos (A - B) + sin² C

Factor:               1 - sin C [cos (A - B) + sin C]

Given:                1 - sin C[cos (A - B) - sin (90° - (A + B))]

Cofunction:       1 - sin C[cos (A - B) - cos (A + B)]

Sum to Product:       1 - sin C [2 sin A · sin B]

                            = 1 - 2 sin A · sin B · sin C

LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C   \checkmark

6 0
3 years ago
At George Washington High School the ratio of graduating seniors that go on to college, as compared to those who do not, is 7 :
soldi70 [24.7K]
First you figure out that the ratio 7:4 is 1.75
Then you do this problem 28 divided by 1.75=16
And 28:16 is the same as 7:4 
Now all you do is add 28+16=44
6 0
3 years ago
−6t−7=17<br><br><br> Who likes Damian Salvatore <br><br> Form vampire diaries
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Answer:

ME! but that's not a real question on top right

6 0
3 years ago
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