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Scilla [17]
3 years ago
9

Solve for x. 1n(5x-3)=2

Mathematics
2 answers:
fredd [130]3 years ago
5 0
So sorry about my handwriting

pav-90 [236]3 years ago
3 0

Answer: x=-1/5n+1

Step-by-step explanation:

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Find the height of a square pyramid that has a volume of 75 ft.³ and a base length of 5 feet
elena-s [515]
I don't know but I guess there is an answer
5 0
3 years ago
How to prove this???
swat32
\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3)  \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 


go to right side now

4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)

use x^3 - y^3 = (x-y)(x^2 + xy + y^2) and x^3 + y^3 = x^2 - xy + y^2

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow  4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad  \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)

so \sin^2 A + \cos^2 A = 1

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side
4 0
3 years ago
Evaluate <br><br> -3+2 / -5+-1 /6+-4 / -2+-7
irinina [24]

Answer: -3+2 = -1 / -5+-1 = -6 / 6+-4 = 2 / -2+-7 = -9

Step-by-step explanation: oof

3 0
3 years ago
Read 2 more answers
What goes in the blanks?
GenaCL600 [577]

Answer:

56/40 + 15/40 = 71/40

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
You flip an unfairly weighted coin 4 times. If the probability of getting a tail is 0. 38.
anygoal [31]

Answer:

0.84306608

Step-by-step explanation:

The probability of an event, e, occurring exactly r times over n trials follows the formula

(n combination r) * p^r * q ^ (n-r)

with p being the probability the event will occur and q being the probability the event will not occur.

I assume you have a calculator/can find one online that can do combinations.

Here, we want to figure out if you get:

- 0 tails

- 1 tail

- 2 tails

If you get 3 or 4 tails, we are getting more than the 2 tails desired

For 0 tails:

- p is the probability a tail will occur = 0.38

- we want it to occur 0 times, so r = 0

- q is (1-p) = 1- 0.38 = 0.62

- we have 4 trials, as we flip it 4 times

(n combination r) * p^r * q ^ (n-r) = (4 combination 0) * (0.38) ^0 * (0.62) ^(4-0) = 1 * (0.38) ^0 * (0.62) ^(4-0) = 0.14776336

For 1 tail:

- p = 0.38, q = 0.62 as with 0 tails

- r = 1, n = 4

(n combination r) * p^r * q ^ (n-r) = (4 combination 1) * (0.38) ^1 * (0.62) ^(4-1) = 4* (0.38) ^1 * (0.62) ^(4-1) = 0.36225856

For 2 tails:

- p = 0.38, q = 0.62 as with 0 tails

- r = 2, n = 4

(n combination r) * p^r * q ^ (n-r) = (4 combination 2) * (0.38) ^1 * (0.62) ^(4-2) = 6* (0.38) ^2 * (0.62) ^(4-2) = 0.33304416

add our 3 probabilities together

0.33304416 + 0.36225856 + 0.14776336 = 0.84306608 as our answer

8 0
2 years ago
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