(trigonometry) write the tangent ratios for angle P and angle Q. if needed reduce fractions

Mathematics

1 answer:

nevsk [136]3 years ago

7 0

Answer:

Step-by-step explanation:

tan P = $\frac{16}{12}$ = $\frac{4}{3}$

tan Q = $\frac{3}{4}$

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Find the inverse of the function x2 + y2 = 4 and the domain of the inverse for 0 ≤ x ≤ 2.

alexandr402 [8]

To get the inverse "relation" of an expression, we first off, do a quick switcharoo of the variables, and then solve for "y", so let's proceed,

$\bf x^2+y^2=4\qquad inverse\implies \boxed{y}^2+\boxed{x}^2=4
\\\\\\
y^2=4-x^2\implies y=\pm\sqrt{4-x^2}$

and yes, the domain for the range 0 ⩽ x <span>⩽ 2, let's get instead the "range" of the original function,

</span> $\bf x^2+y^2=4\implies 0^2+y^2=4\implies y=\pm\sqrt{4}\implies \boxed{y=\pm 2}
\\\\\\
x^2+y^2=4\implies 2^2+y^2=4\implies 4+y^2=4\implies \boxed{y=0}\\\\
-------------------------------\\\\
\stackrel{\textit{range of original}}{-2\le y \le 2}~~=~~\stackrel{\textit{domain of its inverse}}{-2\le x \le 2}$ <span>
</span>

8 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :