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4 years ago

Find the quotient. n^4/n^5

Mathematics

1 answer:

zhenek [66]4 years ago

4 0

Reduce the expression by cancelling the common factors: 1/n

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Solve for x.

Lostsunrise [7]

<h3>Corrected Question: </h3>

Solve for x :

4 - (2x + 4) = 5

1) x = -10

2) x = 5/2

3) x = 1/2

4) x = 6

<h3>Answer: </h3>

➛ 4 - 2x + 4 = 5

➛ 2x + 4 - 4 = 5

➛ 2x + 0 = 5

➛ 2x = 5

➛ x = 5/2

Thus, The value of x is 5/2. 

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2 years ago

How do you solve 9x+2=8x+9

avanturin [10]

Try solving for the variable and get everything to one side

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3 years ago

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For this graph, mark the statements that are true.

ololo11 [35]

Correct Answer are A and D

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3 years ago

If sin theta = (4)/(7), theta in quadrant II, find the exact value of (a) cos theta (b) sin (theta + (pi) / (6) ) (c) cos (the

EleoNora [17]

Answer:

a) $\cos(\theta) = \frac{\sqrt[]{33}}{7}$

b) $\sin(\theta + \frac{\pi}{6})\frac{-3\sqrt[]{11}+4}{14}$

c) $\cos(\theta-\pi)=\frac{\sqrt[]{33}}{7}$

d) $\tan(\theta + \frac{\pi}{4}) = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

Step-by-step explanation:

We will use the following trigonometric identities

$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)$

$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$ $\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$ .

Recall that given a right triangle, the sin(theta) is defined by opposite side/hypotenuse. Since we know that the angle is in quadrant 2, we know that x should be a negative number. We will use pythagoras theorem to find out the value of x. We have that

$x^2+4^2 = 7 ^2$

which implies that $x=-\sqrt[]{49-16} = -\sqrt[]{33}$ . Recall that cos(theta) is defined by adjacent side/hypotenuse. So, we know that the hypotenuse is 7, then

$\cos(\theta) = \frac{-\sqrt[]{33}}{7}$

b)Recall that $\sin(\frac{\pi}{6}) =\frac{1}{2} , \cos(\frac{\pi}{6}) = \frac{\sqrt[]{3}}{2}$ , then using the identity from above, we have that

$\sin(\theta + \frac{\pi}{6}) = \sin(\theta)\cos(\frac{\pi}{6})+\cos(\alpha)\sin(\frac{\pi}{6}) = \frac{4}{7}\frac{1}{2}-\frac{\sqrt[]{33}}{7}\frac{\sqrt[]{3}}{2} = \frac{-3\sqrt[]{11}+4}{14}$

c) Recall that $\sin(\pi)=0, \cos(\pi)=-1$ . Then,

$\cos(\theta-\pi)=\cos(\theta)\cos(\pi)+\sin(\theta)\sin(\pi) = \frac{-\sqrt[]{33}}{7}\cdot(-1) + 0 = \frac{\sqrt[]{33}}{7}$

d) Recall that $\tan(\frac{\pi}{4}) = 1$ and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}=\frac{-4}{\sqrt[]{33}}$ . Then

$\tan(\theta+\frac{\pi}{4}) = \frac{\tan(\theta)+\tan(\frac{\pi}{4})}{1-\tan(\theta)\tan(\frac{\pi}{4})} = \frac{\frac{-4}{\sqrt[]{33}}+1}{1+\frac{4}{\sqrt[]{33}}}$

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3 years ago

PLEASE HELP WITH THIS EASY MATH QUESTION!! it’s down below!!!!!!

ivanzaharov [21]

Answer:

Step-by-step explanation:

x^2 + 2x + 1 = (x+1)^2

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4 years ago

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