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3 years ago

How many different 6-letter arrangements are possible using the letters in ALTUVE?

Mathematics

1 answer:

marshall27 [118]3 years ago

6 0

The answer is 32 because there are 5 in the actual one

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I need to solve for X<br> X=

Gemiola [76]

$\dfrac{2\frac{1}{4}}{\frac{3}{8}}=\dfrac{\frac{15}{16}}{x}\ \ \ \ |\text{cross multiply}\\\\2\dfrac{1}{4}x=\dfrac{3}{8}\cdot\dfrac{15}{16}\\\\\dfrac{2\cdot4+1}{4}x=\dfrac{45}{128}\\\\\dfrac{9}{4}x=\dfrac{45}{128}\ \ \ \ \ |\text{multiply both sides by 4}\\\\9x=\dfrac{45}{32}\ \ \ \ \ |\text{divide both sides by 9}\\\\\boxed{x=\dfrac{5}{32}}$

6 0

3 years ago

find the slope of the line that has the following intercepts: y-intercept: -15; x-intercept: 3. round your answer to 1 decimal p

Readme [11.4K]

Y intercept : (0,-15)
X intercept : (3,0)

Slope : (0 - (-15)) / (3-0) = 15 / 3 = 5

3 0

3 years ago

What is the slope<br> A. Undefined <br> B. 0

Burka [1]

B. is the correct answer love

6 0

3 years ago

Line ab contains the points a(-2,3) and b(4,5). Line ab has a slope that is

gayaneshka [121]

Given two points $A = (A_x, A_y),\ B = (B_x,B_y)$ , the slope of the line passing through the two points is

$m = \dfrac{A_y-B_y}{A_x-B_x}$

So, in your case, you have

$m = \dfrac{3-5}{-2-4} = \dfrac{-2}{-6} = \dfrac{1}{3}$

5 0

3 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

7 0

3 years ago