Find the factors and zeros of 6x^2 +8x-28=2x^2 + 4

Mathematics

1 answer:

Mice21 [21]3 years ago

4 0

Answer:

First solve the equation:

6x^2 + 8x -28 = 2x^2 + 4

=> 6x^2 - 2x^2 + 8x - 28 -4 = 0 => 4x^2 + 8x - 32 = 0

Extract common factor 4:

=> 4[x^2 + 2x - 8] = 0

Now factor the polynomial:

4(x + 4) (x - 2) = 0

=> the solutions are x + 4 = 0 => x = -4, and x - 2 = 0 => x = 2.

So the answer is the option B: 4(x + 4)(x - 2); {-4, 2}

HOPE THIS HELPS! :D

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Bezzdna [24]

Answer:

88°

Step-by-step explanation:

The triangle is an isosceles triangle, so two of the angles are both 46°. We also know that the sum of all the angles in any triangle is 180°, so we can set up the following equation:

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2 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$