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3 years ago

Calculate the area of the figure shown below: https://ibb.co/jSyo5b

Mathematics

1 answer:

Mrrafil [7]3 years ago

7 0

192 don't forget to square root it

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Using the diagram below, translate the figure 3 units to the left and 2 units up. Graph the translated figure and provide the tr

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2 years ago

Solve for X. show your work. (x+2)(x+8)=0

valentinak56 [21]

Step-by-step explanation:

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2 years ago

0.45 plus 9x equals 27

oee [108]

Answer:

x = 2.95

Step-by-step explanation:

Step 1: Convert words into an expression

0.45 plus 9x equals 27

0.45 + 9x = 27

Step 2: Solve for x

0.45 + 9x - 0.45 = 27 - 0.45

9x / 9 = 26.55 / 9

x = 2.95

Answer: x = 2.95

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3 years ago

Read 2 more answers

Cos(A) = ? 13/5 12/5 12/13 5/13

ollegr [7]

Answer:

cosA = $\frac{5}{13}$

Step-by-step explanation:

cosA = $\frac{adjacent}{hypotenuse}$ = $\frac{AC}{AB}$ = $\frac{5}{13}$

6 0

3 years ago

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