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2 years ago

HALP ITS DUE TODAY and somehow im struggling and this is week 3 -_-

Mathematics

1 answer:

Yuliya22 [10]2 years ago

5 0

Answer:

Try to look for the pattern.

As you can see on the first column, the last three numbers line up by +3 (12 to 15 to 18). Proof - 12 + 3 = 15 + 3 = 18.

So you go back by subtracting 3. 12 - 3 = 9 - 3 = 6 - 3 = 3

For the other column, the pattern is +17. Proof - 51 - 34 = 17, 102 - 85 = 17.

So do the same thing as in the first column.

Results:

3 17

6 34

9 51

12 68

15 85

18 102

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Which expressions have a value equal to −22 ?

erica [24]

Answer

-26+4

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2 years ago

PLEASE HELP!

sesenic [268]

Answer:

$36\sqrt{3}$

Step-by-step explanation:

Hi!

Since an equilateral triangle means that every side is equal, our triangle will have 12 on all sides.

To find the height of an equilateral triangle we use $\frac{a\sqrt{3}}{2}$ .

$\frac{12\sqrt{3} }{2} = 6\sqrt{3}$ .

So the height is $6\sqrt{3}$ .

Now we have to solve 12 * $6\sqrt{3}$ ÷ 2.

$12 \cdot 6\sqrt{3} = 72\sqrt{3}$

$72\sqrt{3}\: \div\: 2 = 36\sqrt{3}$ .

Thus, the area of the triangle is $\boxed{36\sqrt{3}}$ .

Hope this helps!

4 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

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

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3 years ago

DINING OUT Trevor and Michelle's restaurant bill comes to $35.50. They are planning to tip the waiter 12%.How much money should

Rzqust [24]

Answer:

4.26

Step-by-step explanation:

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2 years ago

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Molly is designing a dog house for her puppy that has a rectangular base. The length of the base is 7 inches more than its width

Mariana [72]

We assume w is the number of <em>inches</em> of width (as opposed to <em>feet</em> or some other measure). The length is 7 inches more than the width, so is w+7. The area is the product of these dimensions, and the problem statement tells us that area is greater than 375 in².

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3 years ago

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HALP ITS DUE TODAY and somehow im struggling and this is week 3 -_-​

HALP ITS DUE TODAY and somehow im struggling and this is week 3 -_-