$\displaystyle \boxed{(a+b)=a^2+2ab+b^2} \\ \\ (x+3)^2+2(x+3)= \\ x^2+2 \cdot x \cdot 3 +3^2+2x+6 = \\ x^2+6x+9+2x+6= \\ x^2+8x+15= \\ \\ \Delta= b^2-4ac \\ \Delta=8^2-4 \cdot1 \cdot15 \\ \Delta=64-60=4 \\ \\ \\ X_{1,2}= \frac{-b\pm \sqrt{\Delta} }{2a} = \frac{-8\pm2}{2} = \\ \\X1=5 \\ X2=3$

5 0

3 years ago

I'll give a brainliest <br>Find the integers which satisfy the inequality. - 5 < 2n - 1 ≤ 5<br>

ella [17]

Answer:

-1, 0, 1, 2, 3

Step-by-step explanation:

Solve two equations:

-5 < 2n-1 => -4 < 2n => n > -2

and

2n-1 ≤ 5 => 2n ≤ 6 => n ≤ 3

-2 < n ≤ 3

then enumerate the possible values for n

-1, 0, 1, 2, 3

4 0

3 years ago

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Someone please help me

vladimir1956 [14]

Answer:

1. equation: y = 7x, missing = 35

2. equation: y = x - 5, missing = 1

3. equation: y = 1/2x, missing = 2

4. equation: y = x + 4, missing = 15

Step-by-step explanation:

Equation Format: y = mx + b

(m) Slope = $\frac{14-7}{2-1}$

= 7

14 = 7(2) + b

14 = 14 + b

b = 0

y = 7x + 0

Solve for the missing y:

y = 7(5) + 0

y = 35 + 0

y = 35

-Chetan K

3 0

3 years ago

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