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3 years ago

6

A classic counting problem is to determine the number of different ways that the letters of "generally""generally" can be arrang

ed. Find that number.

1 answer:

Dmitrij [34]3 years ago

5 0

The word generally can be arranged more than 5 times.

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How much will 2 3/4 yards of fabric cost at $2.99 per yard

kipiarov [429]

2 3/4 * 2.99 =
11/4 * 2.99 =
32.89/4 =
8.22 <==

or
2 3/4 = 2.75
2.75 * 2.99 = 8.22

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3 years ago

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Hello.

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3 years ago

Find the exact value of cos(theta) for an angle (theta) with tan (theta)= -2/3 and with its terminal side in Quadrant II.

Naya [18.7K]

Answer:

$-\frac{3\sqrt{13}}{13}$ .

Step-by-step explanation:

Since we are in quadrant two, cosine value is negative while sine value is positive.

We are going to use the Pythagorean Identity: $1+\tan^2(\theta)=\sec^2(\theta)$ .

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

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

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

$\frac{13}{9}=\sec^2(\theta)$

$\pm \sqrt{\frac{13}{9}}=\sec(\theta)$

$\pm \frac{\sqrt{13}}{3}=\sec(\theta)$

Since cosine and secant are reciprocals then they will have the same sign as along as they both exist.

$\sec(\theta)=-\frac{\sqrt{13}}{3}$

$\cos(\theta)=-\frac{3}{\sqrt{13}}$ .

I don't see this answer as I'm going to rationalize the denominator.

$\cos(\theta)=-\frac{3}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}}$ .

$\cos(\theta)=-\frac{3\sqrt{13}}{13}$ .

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4 years ago

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Help geometry is so hard for me

Nata [24]

Answer is 5<span>√2 (second choice)
c^2 = 5^2 + 5^2
c^2 = 50
c = </span><span>√50
c = </span>√25 <span>√2
c = 5</span><span>√2</span>

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3 years ago