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Rina8888 [55]

2 years ago

6

Mathematics

1 answer:

ozzi2 years ago

8 0

Answer:

m<SYD = 106.02°. Reason: alternate exterior angles are equal.

Step-by-step explanation:

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Please solve this question

n200080 [17]

Answer:

the answer is b

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Solve the system of equations.

Alenkasestr [34]

Step-by-step explanation:

<u>Step 1: Substitute x from the second equation into the second one</u>

$2x - 9y = 14$

$2(-6y + 7) - 9y = 14$

$(2 * -6y) + (2 * 7) - 9y = 14$

$-12y + 14 - 9y = 14$

$-21y +14 - 14 = 14 - 14$

$-21y/-21 = 0 / -21$

$y = 0$

<u>Step 2: Substitute y into the second equation</u>

$x = -6y + 7$

$x = -6(0) + 7$

$x = 0 + 7$

$x = 7$

Answer: $x = 7, y = 0$

4 0

3 years ago

Read 2 more answers

30 points if you aswer this<img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7Dx%20%5Cfrac%7B3%7D%7B4%7D%20x%5Cfrac%7B4%7D%7B

Licemer1 [7]

Answer:

1/5

Step-by-step explanation:

1/3 x 3/4 x 4/5

1/3 x 3/4 = 3/12

3/12 x 4/5 = 12/60

12/60 = 1/5

8 0

2 years ago

Read 2 more answers

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

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

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

Ming had some candy to give to her five children. She first took three pieces for herself and then evenly divided the rest among

babunello [35]

C-3/5=4

c-3/5*5=4*5 (We do this to get ride 5 so we can simplify to just 3)

c-3=20
+3 +3
c=17

6 0

3 years ago