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3 years ago

14

Show work please I’ll mark brainliest

1 answer:

AveGali [126]3 years ago

6 0

Answer:

2 and 4

Step-by-step explanation:

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HELP QUICKKKKKKKKK NOW I NEED HELPPPPPPPPPPPPPPP PLEASE !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! BRAINLIEST ANYTHING

katrin2010 [14]

Answer:

c) 7

Step-by-step explanation:

square= 3

triangle= 4

3+4+4= 11

3+3+3= 9

3+4= 7

3 0

4 years ago

Read 2 more answers

Rob is travelling to germany he drives for1 1/2 hours wait for 2 3/4 hours flight lasts 2 1/3 hours how long has he travelled pu

DIA [1.3K]

We add all these mixed numbers together:

1 1/2 + 2 3/4 + 2 1/3

It will be easier to convert the mixed numbers into improper fractions:

3/2 + 11/4 + 7/3

Now we have to find the common denominator which is 12:

18/12 + 33/12 + 28/12 = 79/12 = 6 7/12

The answer is 6 7/12 hours.

7 0

4 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

Giving braililist 1+2

vitfil [10]

Answer:

$\huge\boxed{Answer\hookleftarrow}$

$1 + 2 \\ = 3$

3 0

3 years ago

When finding the margin of error for the mean of a normally distributed population from a sample, what is the critical probabili

IceJOKER [234]

<span>To find the margin of error for the mean of a normally distributed population from a sample, the critical probability, $\alpha$ , is obtained by
$p=1-\alpha$
where p is the confidence level.

Assuming a confidence level of 86%, the critical probability is given by
$\alpha =1-0.86=0.14$ </span>

8 0

3 years ago

Read 2 more answers