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3 years ago

6

B is the midpoint of AC. Find the length of the line segment AC IF AB = 48

1 answer:

Gala2k [10]3 years ago

5 0

Answer:

AC=96

Step-by-step explanation:

If B is the midpoint of AC then AB is equal to BC there for you can find AC by adding AB and BC (AB+BC=AC) so in this case it would be 48+48=96

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Numbers 20 - 22 please

Deffense [45]

$(x+5)(x+2)=x^2+2x+5x+10=x^2+7x+10\\\\
(x-1)(2x^2-4x+10)=2x^3-4x^2+10x-2x^2+4x-10=\\
=2x^3-6x^2+14x-10\\\\
(c^2-10)(4c+3)=4x^3+3c^2-40c-30
$

5 0

3 years ago

Z^5=-243i, find the solution to the equation whose argument is strictly between 180 degrees and 270 degrees. Round your answer t

11Alexandr11 [23.1K]

Answer:

The solution is $z = -2.853 - i 0.927$ .

Step-by-step explanation:

Complex power is determined by means of the De Moivre's Theorem, whose expression is:

$z^{n} = r^{n} \cdot (\cos n\theta + i \sin n\theta)$

Where $r$ is the norm of the complex number. In this case, expression can be written as:

$-i 243 = 3^{5} \cdot (\cos 5\theta + i \sin 5\theta)$

The real component must be equal to zero and complex component must be equal to -1. That is to say:

$\cos 5\theta = 0$

$\sin 5\theta = -1$

Possible solutions for each component are, respectively:

Real component

$5\theta = \cos^{-1}0$

$5\theta = \frac{\pi}{2} \pm \pi\cdot j$ , $\forall j \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{10} \pm \frac{\pi}{5} \cdot j$ , $\forall j \in \mathbb{N}_{O}$

Possible solutions: $\frac{11\pi}{10}$ , $\frac{13\pi}{10}$ , $\frac{3\pi}{2}$

Complex component

$5\theta = \sin^{-1}(-1)$

$5\theta = \frac{3\pi}{2} \pm 2\pi \cdot j$ , $\forall j \in \mathbb{N}_{O}$

$\theta = \frac{3\pi}{10} \pm \frac{2\pi}{5} \cdot j$ , $\forall j \in \mathbb{N}_{O}$

Possible solutions: $\frac{11\pi}{10}$ , $\frac{3\pi}{2}$

There is one solution whose argument is strictly between 180 degrees ( $\pi$ ) and 270 degrees ( $1.5\pi$ ).

$z = 3 \cdot \left( \cos \frac{11\pi}{10} + i \sin \frac{11\pi}{10} \right)$

$z = -2.853 - i 0.927$

6 0

3 years ago

Which graph below shows a line that contains ordered pairs that are solutions of the equation shown?

Vinvika [58]

Given:

The equation is:

$y=-x+4$

To find:

The graph of the line that contains ordered pairs that are solutions of the given equation.

Solution:

We need to find the graph of the given equation.

We have,

$y=-x+4$

At $x=0$ , we get

$y=-0+4$

$y=4$

So, the x-intercept of the given equation is at point (0,4).

At $y=0$ , we get

$0=-x+4$

$x=4$

So, the y-intercept of the given equation is at point (4,0).

From the given graphs it is clear that the line in option C has x-intercept at point (0,4) and y-intercept at point (4,0).

Therefore, the correct option is C.

7 0

3 years ago

Btw, the 2nd to last worth says 9th<br> Please solve and explain #1

Amiraneli [1.4K]

81,27, 9, 3, 1 just guess

8 0

3 years ago

The following is a denegerate circle

horrorfan [7]

x+y=2 x^2+y^2=-1 x^2+y<span>^2=1 (x-4)^2+(y-2)^2=0 </span>

8 0

3 years ago