I need help with this

John will paint the four walls of his rectangular basement. The basement measures 10 feet long, 13 feet wide, and 8 feet high an

Svetlanka [38]

Answer:

D. 1,019 square feet

Step-by-step explanation:

10x13x8 - 3x7

3 0

2 years ago

The data below are the number of absences and the final grades of 9 randomly selected students from a literature class. find the

Natasha2012 [34]

I don't know. I apologize.

6 0

3 years ago

A road is inclined at an angle of 5°. After

Scorpion4ik [409]

Answer:

464 feet

Step-by-step explanation:

The Sine of the angle of inclination gives the increase in altitude.

This is because the Sine of an angle is equal to the Opposite side divided by the Hypotenuse side. i.e

Sin Ф = Opposite ÷ Hypotenuse

Sin 5° = O ÷ 5328

0·08716 = O ÷ 5328

Make O the subject and multiply every fraction by 5328

(0·08716/1 × 5328)= (O/5328 × 5328)

464·388 = O

∴ The increase in altitude is 464·388 feet

This is supposed to be to the nearest foot so we round off

464·388 becomes 464 feet

3 0

2 years ago

The first number is 17 the difference between the two numbers on the number line in 9.

Juliette [100K]

Answer:

The second number would be 8 or 16

Step-by-step explanation:

Let x be the unknown number,

Since, first number = 17,

So, the difference between x and 17 = |17 - x |

According to the question,

|17 - x | = 9

17 - x = ± 9

17 - x = 9 or 17 - x = -9

17 - 9 = x or 17 + 9 = x

⇒ x = 8 or x = 26

Hence, the second number would be 8 or 26.

5 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