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fredd [130]
1 year ago
7

Let (-5,-4) be a point on the terminal side of (theta). Find the exact values of cos, csc, and tan.

Mathematics
1 answer:
Rasek [7]1 year ago
3 0

The exact values of the trigonometric identities cos, csc and tan as required in the task content are; -√41/5, -√41/4 and 4/5 respectively.

<h3>What are the exact values of cos, csc and tan as required in the task content?</h3>

It follows from above that the terminal side of the angle theta as described is on the point with coordinates (-5, -4).

Hence, the points spans 5 units leftwards and r units downwards on x and y axis respectively.

Hence, the length of the line that describes the angle by Pythagoras theorem is;

h = √((-4)² + (-5)²)

h = √41.

Hence, it follows from trigonometric identities that;

Cos (theta) = -5/√41.

Csc (theta) = -√41/4

Tan (theta) = 4/5

Read more on trigonometric identities;

brainly.com/question/24496175

#SPJ1

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Noya needs to determine the number of books that will fit into a box. If the box has a length of 18 inches and each book is Star
natima [27]

Answer:

A. 18÷2/9

Step-by-step explanation:

A. 18÷2/9

B. 18×2/9

C. 2/9÷18

D. 9/2÷18

Let

Total length of box = 18 inches

Thickness of each book = 2/9 inches

Number of books that will fit into the box = x

Number of books that will fit into the box = Total length of box ÷ Thickness of each book

x = 18 ÷ 2/9

= 18 × 9/2

= 162/2

= 81 books

The correct answer is A. 18÷2/9

7 0
3 years ago
At Southtown High school the number of students in band is 1 3/4 times the number in orchestra. If 56 students are in orchestra,
NeTakaya

Answer:

98

Step-by-step explanation:

56 x 1 3/4

56 x 1.75 = 98

8 0
2 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

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Pavlova-9 [17]
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In the equation below, identify the constant ; 4(x+1) -7x = -11
castortr0y [4]

Answer:

It will be A because of if you round it it would be -7

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