All categories

3 years ago

Which of the following equations is equivalent to 3x + 4y =15

Mathematics

1 answer:

slavikrds [6]3 years ago

5 0

Answer:

x= −4 /3 y+5

Step-by-step explanation:

You might be interested in

In 2006, out of 100,000 25-year-olds, 99,781 were still alive. The probability of

finlep [7]

Empirical probability

Step-by-step explanation:

5 0

3 years ago

Prove :

Sauron [17]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{\sec\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

We can convert sec(α) to 1 / cos(α):

$\displaystyle \frac{1}{1/\cos\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Multiply both layers of the first fraction by cos(α):

$\displaystyle \frac{\cos\alpha}{1+\cos\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Simplify:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{1-\cos^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{\sin^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Subtract:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Distribute:

$\displaystyle \frac{\cos\alpha-\cos^2\alpha-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Rewrite:

$\displaystyle \frac{(\cos\alpha)-(\cos^2\alpha+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Split:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos^2\alpha+\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{1-\cos^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{(1-\cos\alpha)(1+\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$