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REY [17]
3 years ago
13

I’ll mark you brainlist

Mathematics
2 answers:
Taya2010 [7]3 years ago
7 0

Answer:

c=5

Step-by-step explanation:

igomit [66]3 years ago
3 0

Answer:

C=5

Step-by-step explanation:

23+(-3)=20

20=4c

4*5=20

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HELP QUICK !’ ***What is the area of a circle with a diameter of 126 in?
Vladimir79 [104]

Answer:

49,850.64

Step-by-step explanation:

To find the area of a circle, you must use the formula pi r ^2 (pi r squared).

In this case, we are using 3.14 for pi, so we can add that into the formula.

3.1r^2

We also know the radius, so we can add that in too.

3.14 x 126 ^2

Now solve.

126^2 = 15,876

15,876 x 3.14 = 49,850.64

49,850.64 is the answer.

Hope this helped!

8 0
3 years ago
Solve for u<br> 5+4u=11<br> Simplify your answer as much as possible
r-ruslan [8.4K]

Answer:

u = 1.5

Step-by-step explanation:

5+4u=11

-5        -5

4u = 6

/4      /4

u=1.5

5 0
3 years ago
a total of 695 tickets were sold for the school play. They were either adult or student tickets. There was 55 fewer students tic
FinnZ [79.3K]

Answer:

375 + 320

Step-by-step explanation:

375 adult

320 students

equals 395

4 0
2 years ago
What the answer to x,y, and z
irakobra [83]

You can think of x && y || z as equivalent to: int func (int x, int y, int z) { if (x) { if (y) { return true; } } if (z) { return true; } return false; } Since both x and y are fixed to be non-zero values the first return statement is always hit. thats what i think

6 0
2 years ago
Read 2 more answers
Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.
lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ

Likewise:

\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

m\angle PAC=m\angle PAB+m\angle BAC

Likewise:

m\angle QAB=m\angle QAC+m\angle BAC

Since ∠QAC ≅ ∠PAB:

m\angle PAC=m\angle QAC+m\angle BAC

And by substitution:

m\angle PAC=m\angle QAB

Thus:

\angle PAC\cong \angle QAB

Then by SAS Congruence:

\Delta PAC\cong \Delta BAQ

And by CPCTC:

PC\cong BQ

5 0
3 years ago
Read 2 more answers
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