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sdas [7]
3 years ago
7

If $(2x+5)(x-3)=14$, find the sum of the possible values of $x$.

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0

Answer:

x = 1/4 ± 1/4√233

Step-by-step explanation:

(2x + 5)(x - 3) = 14

~Use FOIL on the left side

2x² - 6x + 5x - 15 = 14

~Combine like terms

2x² - x - 15 = 14

~Subtract 14 to both sides

2x² - x - 29 = 0

~Use the quadratic formula and simplify

x = 1/4 ± 1/4√233

Best of Luck!

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aliya0001 [1]
6x^3+13x^2+16 is the result.
4 0
3 years ago
Solve the Equations
Vikentia [17]

Answer:

1) -2.13

2) 2.57

3) 33

Step-by-step explanation:

1)

3(8 + 5h) = -28  Distribute the 3

24 + 15h = -28  Subtract 24 from both sides of the equation

15h = -32  Divide both sides by 15 and round

h = - 2.13

2)

19 = 7(3n - 5)  Distribute the 7

19 = 21n - 35  Add 35 to both sides of the equation

54 = 21n  Divide both sides of the equation by 21

2.57 = n

3)

6s - 7s = -33  Combine the s's

-s = -33  Multiply both sides by -1

s = 33

5 0
1 year ago
PLEASE HELP FAST
skad [1K]

Answer:

The unit vector u is (-5/√29) i - (2/√29) j

Step-by-step explanation:

* Lets revise the meaning of unit vector

- The unit vector is the vector ÷ the magnitude of the vector

- If the vector w = xi + yj

- Its magnitude IwI = √(x² + y²) ⇒ the length of the vector w

- The unit vector u in the direction of w is u = w/IwI

- The unit vector u = (xi + yj)/√(x² + y²)

- The unit vector u = [x/√(x² + y²)] i + [y/√(x² + y²)] j

* Now lets solve the problem

∵ v = -5i - 2j

∴ IvI = √[(-5)² +(-2)²] = √[25 + 4] = √29

- The unit vector u = v/IvI

∴ u = (-5i - 2j)/√29 ⇒ spilt the terms

∴ u = (-5/√29) i - (2/√29) j

* The unit vector u is (-5/√29) i - (2/√29) j

8 0
2 years ago
Write one word problem using arithmetic sequences and another using geometric sequences.
Levart [38]
Hey There!

If i got this down right,
An = A1 + (n - 1)d
Sn = A1 + A2 + A3 + ... + An
A1 is given by
Sn = n (A1 + An) / 2
For Example

An = A1 + (n - 1)d
= 6 + 3 (n - 1)
= 3 n + 3
n = 50

Hope This Helps!!!
5 0
3 years ago
Read 2 more answers
10) Choose the solution for each linear system.
hodyreva [135]

Answer:

<h2>10.  C. (2, -1)</h2><h2>11.  B. (-2, -4)</h2>

Step-by-step explanation:

10.

\left \{ {{x-2y=4} \atop {4x+2y+6}} \right.

The y values cancel out if you add the equations

5x = 10

x = 2

plug in x

2 - 2y = 4

-2y = 2

y=-1

Answer: C. (2, -1)

11.

\left \{ {{6x-2y=-4} \atop {-4x+3y=-4}} \right.

In order to get terms to cancel out, multiply the top by 3 and bottom by 2

\left \{ {{18x-6y=-12} \atop {-8x+6y=-8}} \right.

The y values cancel out if you add the equations

10x = -20

x = -2

plug in x

6(-2) - 2y = -4

-12 - 2y = -4

-2y = 8

y = -4

Answer: B. (-2, -4)

8 0
2 years ago
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