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Mnenie [13.5K]
3 years ago
6

Simplify 2 to the fifth power over 3 squared all raised to the fourth power. 2 to the twentieth power over 3 to the eighth power

2 to the ninth power over 3 to the eigth power 8 to the fifth power over 12 squared 2 over 3 squared
Mathematics
1 answer:
Murrr4er [49]3 years ago
7 0

2 to the fifth power over 3 squared all raised to the fourth power

\left(\dfrac{2^5}{3^2}\right)^4=\dfrac{(2^5)^4}{(3^2)^4}=\dfrac{2^{5\cdot4}}{3^{2\cdot4}}=\dfrac{2^{20}}{3^8}

Answer: 2 to the twentieth power over 3 to the eighth power

Used:

\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\(a^n)^m=a^{n\cdot m}

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Answer: B

Step-by-step explanation:

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What is the difference between a minor arc and a major arc
Alinara [238K]

Answer:

Minor arcs are linked with smaller than half of a rotation, so minor arcs are linked with angles that are less than 180°. Major arcs are linked with more than half of a rotation, so major arcs are linked with angles greated than 180°.

Step-by-step explanation:

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3 years ago
Zero product property<br> (5y-3) (2y+1)=0<br><br> 2v(3v+4)=0
alexgriva [62]

Answer:

y = 3/5 or y = -1/2

v = 0 oe v = -4/3

Step-by-step explanation:

Zero product property states that if ab = 0, then a = 0 or b = 0.

(5y - 3) (2y + 1) = 0

5y - 3 = 0 or 2y + 1 = 0

y = 3/5 or y = -1/2

2v(3v + 4) = 0

2v = 0 or 3v + 4 = 0

v = 0 oe v = -4/3

8 0
4 years ago
Read 2 more answers
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

8 0
3 years ago
4 less than the product of 9 and a number is 7
Tatiana [17]

Answer:

(9*X)-4=7  :)

Step-by-step explanation:

"*" means times if you didn't know

8 0
3 years ago
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