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4 years ago

Find the complex zeros of x^3+27. write f in factored form

1 answer:

4 0

$x^3+27=0
\\x^3+3^3=0
\\(x+3)(x^2-3x+9)=0
\\x^2-3x+9=0
\\a=1,b=-3,c=9
\\
\\x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac} }{2a} = \frac{-(-3)\pm \sqrt{(-3)^2-4\times1\times9} }{2\times1} =\frac{3\pm \sqrt{-27} }{2} =\frac{3\pm 3\sqrt{3}i }{2} 
\\
\\(x+3)(x-\frac{3+ 3\sqrt{3}i }{2})(x-\frac{3- 3\sqrt{3}i }{2})=0

$

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Give the function for f(x)-5|x+1|+3 for what values of x is f(x)=-12

Lyrx [107]

I’m not sure whether there is a typo in the function but I might have the answer.

If f(x) = 5(x+1)+3 = -12 then you can solve for x.

5(x+1)+3 = -12
5x + 5 + 3 = -12 << expand the brackets
5x + 5 = - 15 << minus 3
x + 1 = -3 << divide everything by 5
x = -4 << minus 1

I don’t know if this helps x Sorry

3 0

3 years ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

3 years ago

What does 8/y + 6 have in common with 8^y + 5?

SpyIntel [72]

Answer:

Answer is d because no is a polynomial

6 0

3 years ago

Read 2 more answers

THIS IS ALGEBRA 2 CAN ANYONE SOLVE IT FOR ME?

Katen [24]

Answer:

(a) 128 ft

(b) 4 s

Step-by-step explanation:

Given information:

h(t) = height in feet
t = time in seconds after the launch

The height of the projectile at launch is the value of h(t) when t = 0 (the y-intercept).

Therefore, from inspection of the graph, the y-intercept is (0, 128).

So the <u>height of the projectile at launch is 128 ft</u>.

The length of time it took for the projectile to land is the time from the beginning (when t = 0) to when the height is 0 (the x-intercept).

From inspection of the graph, the x-intercept is (4, 0)

So the <u>length of time it took for the projectile to land is 4 s</u>.

The vertex is the turning point (minimum/maximum point).

Therefore, from inspection of the graph, the <u>vertex is (1, 144)</u>.

The vertex represents the time and height at which the projectile was at its maximum. So at the time of 1 second, the projectile was at its maximum height of 144 ft.

7 0

2 years ago

Sam typed 100 words in 5 minutes. what is sam’s gwam?

Andru [333]

To find the GWAM (gross words a minute), divide the number of words by the number of minutes so in this case,
100 (words) ÷ 5 (minutes) = 20 GWAM

8 0

4 years ago