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sergij07 [2.7K]
2 years ago
12

B) Solve x ^ 2 + 5x - 14 = 0

Mathematics
2 answers:
uranmaximum [27]2 years ago
6 0

x^2 +5x -14 =0\\\\\implies x^2 +7x -2x -14 =0\\\\\implies x(x+7) -2(x+7)=0\\\\\implies (x-2)(x+7) =0\\\\\implies x =2~~ \text{or}~~ x =-7

Vinil7 [7]2 years ago
6 0

Answer:

x = -7 and x = 2

Step-by-step explanation:

x ^ 2 + 5x - 14 = 0 can be factored as follows:  (x + 7)(x - 2).  Note how 7x - 2x = 5x.

Setting each factor = to 0 and solving for x, we get x = -7 and x = 2.

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Answer:

(60 - 4x)(50 + 2x) = 2,800

Step-by-step explanation:

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3 years ago
I need help with this (simplify 8^3)
sattari [20]

Answer:

512

Step-by-step explanation:

8 = 2^3

Therefore,

8^3  

= (2^3)^3

= 2^(3*3)

= 2^9

= 512

Thus, 8^3 = 512

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5 0
3 years ago
Read 2 more answers
PLEASE HELP
lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function j\left(x\right) \ = \ 4^{x-2}, hence the average rate of change (slope of the blue line) between the x and x+h is

                     \text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\  \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}

                                                            \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}

7 0
1 year ago
HELP HELP HELP HELP PLEASE
Leya [2.2K]

Answer:

-infinity

Step-by-step explanation:

plug in numbers for x that is on the right side of the graph and you will see a trend that the larger x is the bigger y is(technically smaller because y would be negative due to the negative sign outside the square root) as x approaches inifity y approaches negative infinity.

3 0
2 years ago
1.) Create a single variable linear equation that has no solution. Solve the equation algebraically to prove that it does not ha
Mazyrski [523]

9514 1404 393

Answer:

  1. x = x+1
  2. 0 = x+1
  3. x+1 = x+1

Step-by-step explanation:

1. There will be no solution if the equation is a contradiction. Usually, it is something that can be reduced to 0 = 1.

If we choose to make our equation ...

  x = x +1

Subtracting x from both sides of the equation gives ...

  0 = 1

There is no value of the variable that will make this be true.

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2. Something that reduces to x = c will have one solution. One such equation is ...

  0 = x+1

  x = -1 . . . . subtract 1 from both sides

__

3. Something that reduces to x = x will have an infinite number of solutions.

One such equation is ...

  x+1 = x+1

Subtracting 1 from both sides gives ...

  x = x . . . . true for all values of x

3 0
3 years ago
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