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3 years ago

Find the product. (n + 7)(n - 2) n² - 5n - 14 n² + 5n - 14 n² - 5n + 14

2 answers:

6 0

(n+7)(n-2)
=n•n+(-2)•n+7•n+7•(-2)
=n²-2n+7n-14
=n²+5n-14. In conclusion, the product of (n+7) and (n-2) or (n+7)(n-2)=n²+5n-14. Hope it help!

Maslowich3 years ago

3 0

The answer is n^2 + 5n - 14.

$(n+7)(n-2)\\\\n*n+(-2)*n+7*n+7*(-2)\\\\n^{2}-2n+7n-14\\\\n^{2}+5n-14$

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What is the y intercept of the graph A. -2 B. 2 C. -4 D. 4

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Answer:

option C = -4

Step-by-step explanation:

Because the line is in the y-axis of -4

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3 years ago

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2 years ago

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3 years ago

What coordinates on the unit circle are associated with the angle measure?

Varvara68 [4.7K]

The tangent of the given angles are the ratios y to the x coordinate of

the point of the terminal side on the unit circle.

The correct options are;

$\dfrac{34 \cdot \pi}{3} \Longleftrightarrow \underline{ \left(-\dfrac{1}{2}, \ -\dfrac{\sqrt{3} }{2} \right)}$

$-\dfrac{7 \cdot \pi}{4} \Longleftrightarrow \underline{ \left(\dfrac{\sqrt{2} }{2}, \ \dfrac{\sqrt{2} }{2} \right)}$

$210^{\circ} \Longleftrightarrow \underline{ \left(-\dfrac{\sqrt{3} }{2}, \ -\dfrac{1}{2} \right)}$

<h3>How to find the points on the unit circle</h3>

The tangent of an angle is given as follows;

$tan (\theta) = \mathbf{ \dfrac{Opposite}{Adjacent}} = \dfrac{\Delta y}{\Delta x}$

First angle

An angle given is; $\mathbf{\dfrac{34 \cdot \pi}{3}}$

Therefore;

$tan \left(\dfrac{34 \cdot \pi }{3} \right) = \mathbf{ \sqrt{3}}$

The above result can be obtained as follows;

$\sqrt{3} = \mathbf{ \dfrac{-\dfrac{\sqrt{3} }{2} }{-\dfrac{1}{2} }}$

Which is obtained when we have;

$\left( \Delta x, \, \Delta y\right) = \mathbf{\left(-\dfrac{1}{2}, \, -\dfrac{\sqrt{3} }{2} \right)}$

Therefore

The required coordinates is therefore;

$\dfrac{34 \cdot \pi}{3} \Longleftrightarrow \left(-\dfrac{1}{2} , \ -\dfrac{\sqrt{3} }{2} \right)$

Second angle

The angle, $\mathbf{-\dfrac{7 \cdot \pi}{4}}$ , gives; $tan \left(-\dfrac{7 \cdot \pi}{4} \right) = 1$

The above value can be obtained as follows;

$\mathbf{\dfrac{\dfrac{\sqrt{2} }{2} }{\dfrac{\sqrt{2} }{2} }} = 1$

Which gives;

$-\dfrac{7 \cdot \pi}{4} \Longleftrightarrow \left(\dfrac{\sqrt{2} }{2}, \, \dfrac{\sqrt{2} }{2} \right)$

Third angle

The angle 210° gives; tan(210°) = $\mathbf{\frac{1}{\sqrt{3} }}$ , which can be obtained as follows;

$\sqrt{ \dfrac{1}{3} } = \mathbf{\dfrac{-\dfrac{1}{2} }{-\dfrac{\sqrt{3} }{2} }}$

Therefore;

$210^{\circ} \Longleftrightarrow \left(-\dfrac{\sqrt{3} }{2}, \ -\dfrac{1}{2} \right)$

Learn more about the unit circle here:

brainly.com/question/1673530

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2 years ago

A padlock for your gym locker uses a three number sequence to open the lock. if the numbers go from 1 to 28, how many different

aalyn [17]

Answer:

19,656

Step-by-step explanation:

Numbers used in locker = 3

Total numbers available = 28

Repetition is not allowed, so one number can be used only once. The order of number matters in the locker e.g. 123 password is not the same as 231. Since, the order of numbers matter, this is a problem of permutations. We need to find the number of different sequences formed with 28 numbers taken 3 at a time. This can be represented as 28P3

The formula for permutations is:

$^{n}P_{r}=\frac{n!}{(n-r)!}$

For the given case, we will have:

$^{28}P_{3}=\frac{28!}{(28-3)!}\\\\ = \frac{28!}{25!}\\\\ = 19656$

This means, 19,656 different 3 numbered sequences are possible for the locker.

8 0

3 years ago