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-Dominant- [34]
2 years ago
13

Is 9 a solution to the equation: 3x – 5 = 4 + 2x?​

Mathematics
2 answers:
katovenus [111]2 years ago
7 0

Answer:

Yes, 9 will be the answer...

Step-by-step explanation:

3x-5 =4+2x

3x-2x = 4+5 (By transposing method)

x =9

Hope this is correct

Art [367]2 years ago
3 0

Answer:

yeah it is ,

putting the value of x as 9

=> 3x – 5 = 4 + 2x

=> 3×9-5=4+2×9

=> 27-5 = 4+18

=> 22 = 22

✌️

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Move two match sticks to make squares.
Rasek [7]

Answer:

The one in the middle and the one on the right?

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is the number? I did many ways but i didn't found it, please guys help me with it. ​
Advocard [28]

Answer:

A). 15

B). -1.2

Step-by-step explanation:

A). [10(x+3)]-20 = 160

10x+30-20 = 160

10x+10 = 160

10x = 160-10

10x = 150

x = 150/10

x = 15

B). 5{[4(8+(5x))]-9} = 85

5(32+20x+9) = 85

5(32+9+20x) = 85

5(41+20x) = 85

205+100x = 85

100x = 85-205

100x = -120

100x/100 = -120/100

x = -1.2

----------------------------------------------------------------------

HOPE THIS HELPS!!!

3 0
3 years ago
HELP ME PLEASE I REALLY NEED HELP!? :(
Aleksandr [31]
Just answers since long and many simplifiction
it's like doing those multiplication memory things that had like 100 questions but now with algebra fractions

5. fourth or \frac{5x-15}{x^{2}-3x-28}
6. third or\frac{4x+25}{x^{2}+5x+4}
7.the answe ris x+3 since it can be factored out of the 5x+15 tingie (second )
8. fourth or \frac{x^{2}+4x}{4}
9. fourth or \frac{2y+6}{9y-7} times \frac{3}{2}
10. third or \frac{(x+3)(x+8)}{(x-8)(x-8)}
11. second or \frac{x-5}{x-4}
12. third or 11(x+7)=60
5 0
3 years ago
a hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. the c
allochka39001 [22]

According to given conditions, m+n is equal to 409.

Consider the diagram below.

In the hexagon ABCDEF, let

AB = BC = CD = 3;

DE = EF = FA = 5;

Arc BAF is equal to one-third of the circle's circumference.

Hence, ∠BCF = ∠BEF = 60°;

Similarly, ∠CBE = ∠CFE = 60°;

Let the point of intersection of BE and CF be P, BE and AD be Q and CF and AD be R.

∴ Δ EFP and Δ BCP are equilateral, and so Δ PQR is also equilateral.

Also, ∠ BAD and ∠ BED subtend the same arc and so do ∠ ABE and ∠ ADE.

∴ Δ ABQ is similar to Δ EDQ

\frac{AQ}{EQ}  = \frac{BQ}{DQ} = \frac{AB}{ED} = \frac{3}{5}

Also,

\frac{\frac{AD - PQ}{2} }{PQ + 5} = \frac{3}{5}  

and \frac{3 - PQ}{\frac{AD + PQ}{2} }  = \frac{3}{5}

On solving these simultaneous equations, we get AD = 360/49

∴ m + n = 409.

To learn more about similarity of triangles, refer to this link:

brainly.com/question/25882965

#SPJ4

4 0
2 years ago
I really need help on this please
hammer [34]
The answer is letter A
7 0
3 years ago
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