Whats an expression that is equivalent to three over four (5z+ 16).

2 answers:

8 0

<span>(3/4)*(5z+16)
((3/4) * 5z + </span>(3/4) * 16<span>)
3.75z + 12

is that the answer you are looking for?

if you want me to solve for "z" I can but you are asking for an expression not to solve...</span>

Lera25 [3.4K]3 years ago

8 0

<span>(3/4)*(5z+16), ((3/4) * 5z + </span>(3/4) * 16<span>), 3.75z + 12. I think that is right? I'm not 100% sure.</span>

You might be interested in

Please help me out asap! Giving brainliest

irina [24]

The Answer is 3^6

Just for characters

8 0

3 years ago

Read 2 more answers

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 $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .