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aleksley [76]
4 years ago
10

Factor 2x^2+x-3 please

Mathematics
2 answers:
Mamont248 [21]4 years ago
8 0
1
Split the second term into two terms
2{x}^{2}+3x-2x-32x​2​​+3x−2x−3

2 Factor out common terms in the first two terms, then in the last two terms<span>x(2x+3)-(2x+3)<span>x(2x+3)−(2x+3)</span></span>
3 Factor out the common term 2x+32x+3<span><span>(2x+3)(x-1)<span>(2x+3)(x−1)</span></span><span>
</span></span>
RoseWind [281]4 years ago
4 0
(2x + 3)(x - 1)

Is the factorisation.

3 * -1 = -3
3 + (2 * -1) = 1
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Зу - 2x = -2<br> у+ 2x = 8
Andrew [12]

Answer:

(3.25, 1.5)

Step-by-step explanation:

<em><u>First, turn the 2nd equation into y=mx+b form:</u></em>

у+ 2x = 8

<u>  -2x     -2x</u>

y  =  -2x + 8

<u><em>Next Substitute:</em></u>

З(-2x+8) - 2x = -2

-6x + 24 - 2x = -2

-8x + 24 = -2

<u><em>Then simplify:</em></u>

-8x + 24 = -2

<u>      -24     -24</u>

-8x       = -26

x = 3.25

<em><u>Lastly, substitute x into any one of the original equations:</u></em>

y + 2 (3.25) =8

y + 6.5 = 8

<u>   - 6.5   - 6.5</u>

y  =  1.5

please give me brainliest i worked hard on this :,)

and its 1000% correct, i checked it i promise

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3 years ago
Find the solution to the system. Write your solution as an ordered pair (x,y) with no spaces, no solution, or infinitely many. X
chubhunter [2.5K]

Answer:

(15, 11).

Step-by-step explanation:

Given the system of equation;

X=2y-7 ..... 1

4x+y=71 ....2

Substitute 1 into2;

4(2y-7) + y = 71

8y - 28 + y = 71

9y - 28 = 71

9y = 71 + 28

9y = 99

y = 11

Substitute y = 11 into 1;

x = 2y - 7

x = 2(11)-7

x = 22-7

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hence the solution to the equation is (15, 11).

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Step-by-step explanation:

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\int\limits^{3x}_x { \frac{1}{t} } \, dt
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3 years ago
Please answer thank you
Nimfa-mama [501]

Answer:

Yes, (0,4) is a solution

Step-by-step explanation:

We have to plug in 0 in x and 4 in y IN BOTH THE INEQUALITIES.

IF BOTH ARE TRUE, then the system of inequalities is TRUE.

<u>Let's check:</u>

y ≤ -3x+4

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Is 4 less than OR equal to 4? Yes. THis is satisfied.

<u>Now, checking 2nd one:</u>

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