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Rudiy27
4 years ago
12

Is it an easier way to find previous answers besides scrolling down so much stuff? Thanks

Mathematics
1 answer:
Nimfa-mama [501]4 years ago
7 0
Yes on the top where it says "what is your question?" type in your question you need...
Hope this has helped you! 
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Heyy! i’ll give brainliest please help
almond37 [142]

Answer:

d. Atlantic Ocean

8 0
3 years ago
Read 2 more answers
3+p/2=6 How do I solve this?
Diano4ka-milaya [45]

\boxed{\underline{\bf \: ANSWER}}

\sf\frac { 3 + p } { 2 } = 6 \\

Multiply both sides by 2.

\sf \: 3+p=6\times 2

Multiply 6 and 2 to get 12.

\sf \: 3+p=12

Subtract 3 from both sides.

\sf \: p=12-3

Subtract 3 from 12 to get 9.

\boxed{\bf \: p=9}

____

Hope it helps.

RainbowSalt2222

5 0
3 years ago
Please assist with this question!!!!!!!!!! No links or your account will be banned
andriy [413]

Answer:

8/15

Step-by-step explanation:

4 0
2 years ago
Write a two-column proof. Given: line BD bisects angle CBE. Prove: angle ABD approximately equal to angle FBD.
ale4655 [162]

Answer:

The proof is derived from the summarily following equations;

∠FBE + ∠EBD = ∠CBA + ∠CBD

∠FBE + ∠EBD = ∠FBD

∠CBA + ∠CBD = ∠ABD

Therefore;

∠ABD ≅ ∠FBD

Step-by-step explanation:

The two column proof is given as follows;

Statement {}                                       Reason          

\underset{BD}{\rightarrow} bisects ∠CBE {}                            Given

Therefore;

∠EBD ≅ ∠CBD  {}                              Definition of angle bisector

∠FBE ≅ ∠CBA {}                                Vertically opposite angles are congruent

Therefore, we have;

∠FBE + ∠EBD = ∠CBA + ∠CBD {}     Transitive property

∠FBE + ∠EBD = ∠FBD {}                    Angle addition postulate

∠CBA + ∠CBD = ∠ABD {}                   Angle addition postulate

Therefore;

∠ABD ≅ ∠FBD        {}                          Transitive property.

3 0
3 years ago
8n - (2n+11 - 54)<br>simplify
Serjik [45]

Answer:

8n - (2n + 11 - 54) =

Step-by-step explanation:

6n + 43, I used a calculator !

5 0
3 years ago
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