Cos(theta) has positive values in quadrants I and IV and it has negative value in quadrants II and III. To know that you don't have to remember this but to imagine xy coordinate system. In it you draw a circle around the center.

Once you do that for any given angle you project the point which represents crossing of side that defines angle and circle. Now project that dot on x-axis. If projection is on positive side of x-axis, cos of that angle is positive and if it is on negative side it is negative.

7 0

4 years ago

Someone please help meeee 546cm2

Nimfa-mama [501]

12 cm
Explanation
:&:837:747

5 0

4 years ago

Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. 5x − 4 x(x2 + 7

Nikitich [7]

Answer:

$\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}$

Step-by-step explanation:

Given the expression $\frac{5x-4}{x(x^2+7)^2}$ , we are to re-write the expression in form of a partial fraction.

Before we write in form of a partial fraction, we need to note the expression at the denominator. Since the expression in parenthesis is a quadratic equation, the equivalent numerator must be a linear expression.

Also the quadratic equation is a repeated form since it is squared. This means that we are to repeat the quadratic equation twice when writing as a partial fraction.

$\frac{5x-4}{x(x^2+7)^2} = \frac{A}{x} + \frac{Bx+C}{x^2+7} + \frac{Dx+E}{(x^2+7)^2}$

From the above partial fraction, it can be seen that x² + 7 in parenthesis was repeated twice and their equivalent expressions at the numerator are both linear i.e Bx+E and Dx+ E where A, B, C, D and E are the unknown constant.

4 0

3 years ago