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dem82 [27]
3 years ago
7

Find the solutions of the congruence 12x2 + 25x ≡ 10 (mod 11). [Hint: Show the congruence is equivalence to the congruence 12x2

+ 25x + 12 ≡ 0 (mod 11). Factor the left-hand side of the congruence; show that a solution of the quadratic congruence is a solution of one of two different linear congruences.]
Mathematics
1 answer:
Vlada [557]3 years ago
5 0

Answer:23.89x67

Step-by-step explanation:

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Please help with photo below
gavmur [86]

Answer:

No

Step-by-step explanation:

like terms are terms which have the same variable , that is

10x and - 5x are like terms, they both have the same variable x

6 0
2 years ago
What is... X/2 - 13 = 1
kow [346]
X ÷ 2 - 13 = 1
Write the division as a fraction
\frac{1}{2}x - 13 = 1
Multiple both sides of the equation by 2
2 \times  \frac{1}{2}x  \times  - 2 \times 13 = 12
Reduce the number with greatest common divisor 2
x - 2 \times 3 = 2
Multiply the numbers
- 2 \times 13
Then u will have
x - 26 = 2
Move constant to the right by adding its opposite sides
x - 26 + 26 = 2 + 26
Eliminate the opposites
- 26 + 26
Then you have
x = 2 + 26
Add the numbers
x = 28

5 0
3 years ago
14+16+49+10<br><br>Hint: Add what will get a tens number first!
Leviafan [203]
<span>89 is the correct answer.

</span>
8 0
3 years ago
Read 2 more answers
Can some one please help and please hurry​
fenix001 [56]
Hi don’t know but have a wonderful day
8 0
3 years ago
Calculate the partial sum S for the sequence 243,81,27,....
Llana [10]

Answer:363


Step-by-step explanation:

We have to find partial sum for the sequence 243 , 81 , 27 ..... up to 5 terms(S5 given)

The sequence actually is 3^5,3^4.,3^3.....

Therefore first 5 terms are 1) 3^5 I.e. 243

2) 3^4 I.e. 81

3) 3^3 I.e. 27

4)) 3^2 I.e. 9

5) 3^1 I.e. 3

Adding all those no. we get partial sum of first 5 no. of the sequence

So, 243 + 81 + 27 + 9 + 3

= 363

Hope it helps!!!

4 0
3 years ago
Read 2 more answers
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