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3 years ago

Someone Please Help Me With This n _ = 1 6 3

Mathematics

1 answer:

Ivahew [28]3 years ago

3 0

Answer:

n=48

Step-by-step explanation:

_ = 1 6

Multiply each side by 3

n/3 * 3 = 16*3

n = 48

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Math:

Dahasolnce [82]

Answer:

a) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$ , b) $x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$ , $x_{2} = \frac{5\pi}{3}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

Step-by-step explanation:

a) The equation must be rearranged into a form with one fundamental trigonometric function first:

$\sqrt{3}\cdot \csc x - 2 = 0$

$\sqrt{3} \cdot \left(\frac{1}{\sin x} \right) - 2 = 0$

$\sqrt{3} - 2\cdot \sin x = 0$

$\sin x = \frac{\sqrt{3}}{2}$

$x = \sin^{-1} \frac{\sqrt{3}}{2}$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

$x_{2} = \frac{5\pi}{6}\pm 2\pi\cdot i$ , $\forall i \in \mathbb{N}_{O}$

b) The equation must be simplified first:

$\cos x + 1 = - \cos x$

$2\cdot \cos x = -1$

$\cos x = -\frac{1}{2}$

$x = \cos^{-1} \left(-\frac{1}{2} \right)$

Value of x is contained in the following sets of solutions:

$x_{1} = \frac{\pi}{3}\pm 2\pi \cdot i$ , $\forall i \in \mathbb{N}_{O}$

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

7 0

3 years ago

Solve for y in terms of x <br><br> -8x + 4y = -16

solniwko [45]

Answer:

y=2x-4

Step-by-step explanation:

well first you have to move the -8x to the other side where the -16 is. by doing so, the -8x will turn into a POSTIVE 8x. than your problem should look like (4y=8x-16). now that your problem is like that, you want "y" to be alone. so, you divide the 4 by every number. (so you would do 4y/4, 8x/4, and -16/4). than your final product would be: y=2x-4

4 0

3 years ago

Read 2 more answers

Georgetown and Franklin are 9.7 in apart on a map that has a scale of 1.1 in : 15 mi. How far apart are the real cities?

lina2011 [118]

$\bf \begin{array}{ccllll}
inches&miles\\
\textendash\textendash\textendash\textendash\textendash\textendash&\textendash\textendash\textendash\textendash\textendash\textendash\\
1.1&15\\
9.7&x
\end{array}\implies \cfrac{1.1}{9.7}=\cfrac{15}{x}\impliedby \textit{solve for "x"}$