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AleksAgata [21]
2 years ago
11

PLEASE HELP!!! TRIG, Theres a picture!!

Mathematics
1 answer:
BartSMP [9]2 years ago
3 0

Answer:

<h2>The answer is option C</h2>

Step-by-step explanation:

3(  { \tan }^{2} \theta -    { \sec }^{2}  \theta)

Using trigonometric identities

That's

1 + { \tan }^{2} \theta  =   { \sec }^{2}  \theta

Rewrite the expression

That's

3({ \tan }^{2} \theta - (1 + { \tan }^{2} \theta)) \\

Simplify

We have

3({ \tan }^{2} \theta - 1 - { \tan }^{2} \theta) \\ 3({ \tan }^{2} \theta - { \tan }^{2} \theta - 1)

So we have

3( - 1)

We have the final answer as

<h2>- 3</h2>

Hope this helps you

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If BC = 5, DE = 9, and AB = 4, what is the length of AD?<br><br><br> 7<br> 8<br> 7.2<br> 7.5
steposvetlana [31]

There are two triangles here: ABC and ADF. These triangles are similar, and thus proportional. So, we can set up a proportion between the left side of each triangle and the base of each triangle.

AB / AD = BC / DE

4 / (4 + BD) = 5 / 9

---AD is made up of AB and DB. We know AB, but not BD, so we need to find its value to find the length of AD.

5(4 + BD) = 9 x 4

20 + 5BD = 36

5BD = 16

BD = 3.2

AD = AB + BD

AD = 4 + 3.2

AD = 7.2

Hope this helps!

8 0
3 years ago
Which expression is a fourth root of -1+isqrt3?
aleksklad [387]

Answer:

Step-by-step explanation:

\sf n^{th} roots of a complex number is given by DeMoivre's formula.

   \sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}

Here, k lies between 0 and (n -1) ; n is the exponent.

\sf -1 + i\sqrt{3}

a = -1 and b = √3

\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}

\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}

                   \sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})

                   \sf = \dfrac{-\pi }{3}

n = 4

For k = 0,

          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

4 0
1 year ago
The workers in a senate office said that they received about 700 phone calls from
prohojiy [21]

Answer:

672

Step-by-step explanation:

anything lower then 5 or in this case 50 would round down so 649 rounds to 600 anything higher would round up 672 to 709 763 to 800 and 751 to 800

7 0
2 years ago
Graph the follow in equation
AURORKA [14]
13 and 2 are numbers you could use in graph☺
8 0
2 years ago
Read 2 more answers
I need help.<br><br> someone help please!
Vikentia [17]
Sorry, I think there is missing context?
4 0
3 years ago
Read 2 more answers
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