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prisoha [69]
3 years ago
11

Multiply (FOIL) ( x - 3)( 2x + 5) (3x + 7)(2x - 1)

Mathematics
1 answer:
Troyanec [42]3 years ago
7 0

(x-3)(2x+5)=2x^2+5x-6x-15=\boxed{2x^2-x-15}\\\\(3x+7)(2x-1)=6x^2-3x+14x-7=\boxed{6x^2+11x-7}

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6a + 4 pls I will give brainlist if ur answer is right
Ad libitum [116K]

Answer:

Step-by-step explanation:

6a + 4

Common factor

6a + 4 - using common factor awnser = solution down below:

2(3a + 2)

8 0
3 years ago
The roots of \[z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\]are $\text{cis } \theta_1$, $\text{cis } \theta_2$, $\dots$, $\te
stepan [7]

Decoding the LaTeX that didn't render, we seek sum of the angles of the seventh roots of

- \frac 1 {\sqrt 2} - \frac i {\sqrt{2}}

That's on the unit circle, 45 degrees into the third quadrant, aka 225 degrees.

The seventh roots will all be separated by 360/7, around 51 degrees. The first seventh root has

\theta_1 = 225^\circ / 7

That's around 32 degrees.

The next angle is

\theta_2 = \frac{1}{7}( 225^\circ +  360^\circ)

The next one is

\theta_3 = \frac{1}{7}( 225^\circ +  720^\circ)

and in general

\theta_n = \frac{1}{7}( 225^\circ +  (n-1)360^\circ) = \frac 1 7(-135^\circ + 360^\circ n)

S = \displaystyle \sum_{n=1}^7 \theta_n = \frac{1}{7} \sum_{n=1}^7 -135^\circ +  \frac{360}{7} \sum_{n=1}^7 n

The first sum is just -135° since it's one seventh of the sum of seven -135s.

We have 1+2+3+4+5+6+7 = (1+7)+(2+6)+(3+5) + 4 = 28 so

S = -135^\circ + 360^\circ (\frac{28} 7) = 4(360)-135 = 1305^\circ

If I didn't screw it up, that means the answer is

Answer: 1305°


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