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uranmaximum [27]

3 years ago

15

Problema 3

1 answer:

Serjik [45]3 years ago

4 0

espero y te ayude, cualquier cosa escríbeme

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Check all that apply. If csc =13/5 , then:

nirvana33 [79]

Answer:

The answer are the options B and C

$sin(x)=\frac{5}{13}$

$tan(x)=\frac{5}{12}$

Step-by-step explanation:

we know that

$csc(x)=\frac{1}{sin(x)}$

In this problem we have

$csc(x)=\frac{13}{5}$

substitute

$\frac{13}{5}=\frac{1}{sin(x)}$

solve for sin(x)

$sin(x)=\frac{5}{13}$

$cos^{2}(x)+sin^{2}(x)=1$

substitute the value of sin(x) and solve for cos(x)

$cos^{2}(x)+(\frac{5}{13})^{2}=1$

$cos^{2}(x)=1-(\frac{5}{13})^{2}$

$cos^{2}(x)=(\frac{13^{2}-5^{2}}{13^{2}})$

$cos^{2}(x)=(\frac{144}{13^{2}})$

$cos^{2}(x)=(\frac{12^{2}}{13^{2}})$

$cos(x)=(\frac{12}{13})$

The function tangent is equal to

$tan(x)=\frac{sin(x)}{cos(x)}$

substitute the values

$tan(x)=\frac{(5/13)}{(12/13)}=5/12$

8 0

3 years ago

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The right answer can you help me plz

MArishka [77]

sooooooo....whats the question???

6 0

3 years ago

Using the Pythagorean theorem, find the length of c if a = 14 * the square root of 3 and b = 34

ANEK [815]

If c is the hypotenuse, we have

$c=\sqrt{a^2+b^2}=\sqrt{(14\sqrt{3})^2+34^2}=\sqrt{588+1156}=\sqrt{1744}$

If instead b is the hypothenuse, you have

$c=\sqrt{b^2-a^2}=\sqrt{1156-588}=\sqrt{568}$

5 0

3 years ago

Anyone? please help asap i really want to finish this

Nesterboy [21]

Answer:

Question 3: $b \geq 2$

Question 4: $r < -5$

Question 5: $y \leq -6$

Step-by-step explanation:

Question 3:

$b + 2 \geq 4\\b + 2 -2 \geq 4 - 2\\b \geq 2\\$

Question 4:

$13 > 18 + r\\13 - 18 > 18 -18 + r\\\\-5 > r\\ r < -5$ sign flips when r switches sides

Question 5:

$3y \leq 2y - 6\\3y - 2y \leq 2y - 2y - 6\\\\y \leq - 6\\$

8 0

2 years ago

Bryan likes to do math problems everyday for practice. He can complete 3 problems in 9 minutes. At that rate, how many problems

mixer [17]

Answer:

2 problems in 6 minutes

Step-by-step explanation:

if 1 problem is three minutes 2 should be six

7 0

3 years ago

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