What is the value of Q?

Mathematics

1 answer:

butalik [34]2 years ago

4 0

Answer:

52 degrees

Step-by-step explanation:

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Factor to find the zeros of the function defined by the quadratic expression. −7x2 − 91x − 280

Mariulka [41]

There are no zeroes for an expression, only for equations.
Assuming equation to be
−7x^2 − 91x − 280=0
-7(x^2+13x+40)=0
-7(x+8)(x+5)=0
by the zero product properties,
x+8=0 => x=-8
or
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Find all solutions of the equation in the interval [0,2pi)

irinina [24]

Answer:

{π/4, 5π/4}

Step-by-step explanation:

Tan theta -1=0 could be rewritten as tan Ф = 1. The tangent function is 1 at Ф = π/4. As the period of the tangent function is π,

tan Ф = 1 will be true for Ф = π/4 + π, or (5/4)π.

The solution set is {π/4, 5π/4}.

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2 years ago

Is the following shape a rectangle? how do you know?

Usimov [2.4K]

Slope of AB = 1/3 and slope of BC = -3 so these 2 lines are perpendicular
The same is true for all the other adjjacent pairs of lines.
Oppoitse lines are also paralllel ( slope of AB = 1/3 and slope of DC = 1/3) and other pair are both of slope -3.

So Its C

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2 years ago

Please provide answer and work to show how you got the answer!!

yan [13]

X=8
Here my
Work message me if u have questions

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Read 2 more answers

Please Help! This is a trigonometry question.

liraira [26]

$\large\begin{array}{l} \textsf{From the picture, we get}\\\\ \mathsf{tan\,\theta=\dfrac{2}{3}}\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{2}{3}}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\qquad\mathsf{(i)} \end{array}$

$\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}$

$\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}$

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$\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}$

________

$\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}$

$\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}$

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$\large\textsf{I hope it helps.}$

Tags: trigonometry trig function cosecant csc double angle identity geometry

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3 years ago