What is 6.33 percent as a decimal

Mathematics

2 answers:

Harrizon [31]4 years ago

5 0

I'm pretty sure the answer is .0633 but I'm not sure

denis23 [38]4 years ago

4 0

6.33 Percent As A Decimal = 0.0633

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Find the equation of the line through point (3,−3) and parallel to y=3x−4. Use a forward slash (i.e. "/") for fractions (e.g. 1/

exis [7]

Answer:

y =3x - 12

Step-by-step explanation:

I'm fairly certain this is the answer.

Consider the point (3,-3). In terms of variables this is written (x,y).

Remember that the equation for slope is y = mx + b.

In that equation, "b" is your y-intercept, and "m" is your slope.

If 2 lines are parallel, they have the same slope but different y-intercepts. So you can use the slope from the equation y=3x-4 to find the parallel line. You can also plug in the values 3 and -3 to solve for b, our y-intercept.

If we set up the equation, we have -3 = 3(3) + b.

Now solve for b.

-3 = 9 + b

Combine like terms, so that -12 = b, and you have your y-intercept.

To create the equation, take out the coordinate pair that the line passes through, and only have the SLOPE and the Y-INTERCEPT showing in the equation.

Therefore, y = 3x - 12

5 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .