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2 years ago

An acute angle theta is in a right triangle with cos theta = 5/6. What is the value of sec theta?

Mathematics

1 answer:

pochemuha2 years ago

5 0

Answer:

sec=1/cos

Step-by-step explanation:6/5

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How do you solve this?

Katarina [22]

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] ~\dotfill\\\\ (x-6)^2+(y+5)^2=16\implies [x-\stackrel{h}{6}]^2+[y-(\stackrel{k}{-5})]^2=\stackrel{r}{4^2}~\hfill \begin{cases} \stackrel{center}{(6,-5)}\\ \stackrel{radius}{4} \end{cases}$

Check the picture below.

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3 years ago

Read 2 more answers

Could you please answer Number 2? This is for geometry

katrin [286]

Answer: 108 degrees

============================================================

Explanation:

Angle ABC is given to be 90 degrees. By the converse of Thales Theorem, we know that AC is a diameter of the circle.
If we draw a line from A to C, it will pass through the center of the circle.

Therefore, arc AC is 180 degrees as this is half the distance around the circle.

minor arc AD = 72
minor arc CD = x
(minor arc AD) + (minor arc CD) = arc AC
72+x = 180
72+x-72 = 180-72
x = 108

minor arc CD is 108 degrees

7 0

3 years ago

Which relation is a function

Anon25 [30]

$\{(\boxed{2};\ 3);\ (1;\ 5);\ (\boxed{2};\ 7)\}-NO\\\\\{(-1;\ 5);\ (-2;\ 6);\ (-3;\ 7)\}-YES\\\\\{(\boxed{11};\ 9);\ (\boxed{11};\ 5);\ (9;\ 3)\}-NO\\\\\{(\boxed{3};\ 8);\ (0;\ 8);\ (\boxed{3};\ -2)\}-NO$

3 0

2 years ago

-12 - 6p - (-2) combining like terms with negative coefficients

vlada-n [284]

- 12 - 6p - (-2) =

= - 12 - 6p + 2

= - 10 - 6p

= - 2(5 + 3p)

4 0

2 years ago

The function below models the voltage, in volts, of a certain alternating current after x seconds, where A and b are positive co

Katarina [22]

First of all, we can just ignore A, it has no effect but to vertically stretch our cosine.

If it was only $f(x)=cosx$ , the function would be invertible as long as it's confined between $0$ and $\pi$ . Now, the argument of our cosine is not $x$ but $bx$ . It means that it won't stop at $\pi$ , but at $\frac{\pi}{b}$ .

Another way to think about it, "what should i replace x with so I get $\pi$ inside the cosine?"

4 0

2 years ago