Triangle PQR has vertices $P(-2, 6), \ Q(-8, 4), and\ R(1, -2).$ It is translated according to the rule $(x, y)\ -> \ (x\ - \ 2, y\ - 16)$ . What is the y-value of P'?

Now coming to the solution as already mention in the question

The translated rule is $(x, y)\ -> (x - 2, \ y -16).$

Now calculated the vertices P value according to the rule of translated

$P(-2, 6)\\Now \ apply \ the\ translated\ rule \ in\ P\ vertices\\P(-2, 6)->P1(-2-2,\ 6-16)\\P1->(-4,-10)$

So -10 is the value of y in P vertices .

6 0

3 years ago

Read 2 more answers

. Find the simple interest on a $2,219.00 principal, deposited for 6 years at a

koban [17]

Answer:

SI=PTR/100

SI=2129*6*1.91/100

SI=24398.34/100

SI=243.9834$

7 0

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