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3 years ago

Which values are solutions of the quadratic equation 0 = (x + 3)2 – 4? Check all that apply.

Mathematics

1 answer:

larisa86 [58]3 years ago

5 0

Answer:

The answer is -1.

Step-by-step explanation:

(-1 + 3)2-4

(2)2-4

4-4

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Find the derivative

zalisa [80]

First use the chain rule; take $y=\dfrac{x+5}{x^2+3}$ . Then

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dy}\cdot\dfrac{\mathrm dy}{\mathrm dx}$

By the power rule,

$f(x)=y^2\implies\dfrac{\mathrm df}{\mathrm dy}=2y=\dfrac{2(x+5)}{x^2+3}$

By the quotient rule,

$y=\dfrac{x+5}{x^2+3}\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)\frac{\mathrm d(x+5)}{\mathrm dx}-(x+5)\frac{\mathrm d(x^2+3)}{\mathrm dx}}{(x^2+3)^2}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{(x^2+3)-(x+5)(2x)}{(x^2+3)^2}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3-10x-x^2}{(x^2+3)^2}$

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)}{x^2+3}\cdot\dfrac{3-10x-x^2}{(x^2+3)^2}$

$\implies\dfrac{\mathrm df}{\mathrm dx}=\dfrac{2(x+5)(3-10x-x^2)}{(x^2+3)^3}$

6 0

3 years ago

Belle bought 18 seeds in order to plant an herb garden for her grandma. Of the seeds she bought, 10 were parsley seeds.

Harrizon [31]

Answer:

0.3431

Step-by-step explanation:

Here, it can work well to consider the seeds from the group of 18 that are NOT selected to be part of the group of 15 that are planted.

There are 18C3 = 816 ways to choose 3 seeds from 18 NOT to plant.

We are interested in the number of ways exactly one of the 10 parsley seeds can be chosen NOT to plant. For each of the 10C1 = 10 ways we can ignore exactly 1 parsley seed, there are also 8C2 = 28 ways to ignore two non-parsley seeds from the 8 that are non-parsley seeds.

That is, there are 10×28 = 280 ways to choose to ignore 1 parsley seed and 2 non-parsley seeds.

So, the probability of interest is 280/816 ≈ 0.3431.

___

The notation nCk is used to represent the expression n!/(k!(n-k)!), the number of ways k objects can be chosen from a group of n. It can be pronounced "n choose k".

6 0

3 years ago

Which function is the same as y = 3 cosine (2 (x startfraction pi over 2 endfraction)) minus 2? y = 3 sine (2 (x startfraction p

kirza4 [7]

The function which is same as the function y = 3cos(2(x +π/2)) -2 is: Option A: y= 3sin(2(x + π/4)) - 2

<h3>How to convert sine of an angle to some angle of cosine?</h3>

We can use the fact that:

$\sin(\theta) = \cos(\pi/2 - \theta)\\\sin(\theta + \pi/2) = -\cos(\theta)\\\cos(\theta + \pi/2) = \sin(\theta)$

to convert the sine to cosine.

<h3>Which trigonometric functions are positive in which quadrant?</h3>

In first quadrant (0 < θ < π/2), all six trigonometric functions are positive.
In second quadrant(π/2 < θ < π), only sin and cosec are positive.
In the third quadrant (π < θ < 3π/2), only tangent and cotangent are positive.
In fourth (3π/2 < θ < 2π = 0), only cos and sec are positive.

(this all positive negative refers to the fact that if you use given angle as input to these functions, then what sign will these functions will evaluate based on in which quadrant does the given angle lies.)

Here, the given function is:

$y= 3\cos(2(x + \pi/2)) - 2$

The options are:

$y= 3\sin(2(x + \pi/4)) - 2$
$y= -3\sin(2(x + \pi/4)) - 2$
$y= 3\cos(2(x + \pi/4)) - 2$
$y= -3\cos(2(x + \pi/2)) - 2$

Checking all the options one by one:

Option 1: $y= 3\sin(2(x + \pi/4)) - 2$

$y= 3\sin(2(x + \pi/4)) - 2\\y= 3\sin (2x + \pi/2) -2\\y = -3\cos(2x) -2\\y = 3\cos(2x + \pi) -2\\y = 3\cos(2(x+ \pi/2)) -2$

(the last second step was the use of the fact that cos flips its sign after pi radian increment in its input)
Thus, this option is same as the given function.

Option 2: $y= -3\sin(2(x + \pi/4)) - 2$

This option if would be true, then from option 1 and this option, we'd get:
$-3\sin(2(x + \pi/4)) - 2= -3\sin(2(x + \pi/4)) - 2\\2(3\sin(2(x + \pi/4))) = 0\\\sin(2(x + \pi/4) = 0$