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3 years ago

G(x)=(cosθsinθ)^4 what's the differential

Mathematics

1 answer:

natulia [17]3 years ago

4 0

Answer:

sin²2θ. (cos θ sin θ). cos 2θ

Step-by-step explanation:

finding g'(x)

g'(x)

(x^n)' = nx^(n -1)

= 4 (cosθsinθ)³ . { cosθ. (sinθ)' + sinθ. (cosθ)' }

(cosθ)' = - sinθ
(sinθ)' = cosθ

= 4 (cosθsinθ)³ { cosθ. cos θ + sinθ.(-sin θ)}

= 4 (cosθsinθ)³{ cos²θ - sin²θ}

cos²θ - sin²θ = cos 2θ
2sinθ cosθ = sin 2θ

= (4 cosθ sinθ)². (cosθ sinθ). { cos²θ - sin²θ}

= sin²2θ. (cos θ sin θ). cos 2θ

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Solve using the elimination method x + 5y = 26 - X+ 7y = 22

S_A_V [24]

Answer:

$x=6\\y=4$

Step-by-step explanation:

Elimination method:

$x+5y=26$

$-x+7y=22$

Add these equations to eliminate x:

$12y=48$

Then solve $12y=48$ for y:

$12y=48$

$y=48/12$

$y=4$

Write down an original equation:

$x+5y=26$

Substitute 4 for y in $x+5y=26$ :

$x+5(4)=26$

$x+20=26$

$x=26-20$

$x=6$

{ $x=6$ and $y=4$ } ⇒ $(6,4)$

hope this helps...

5 0

3 years ago

Read 2 more answers

Its a graph problem

OLga [1]

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8 0

3 years ago

Evaluate each finite series for the specified number of terms. 1+2+4+...;n=5

zaharov [31]

Answer:

Step-by-step explanation:

The series are given as geometric series because these terms have common ratio and not common difference.

Our common ratio is 2 because:

1*2 = 2

2*2 = 4

The summation formula for geometric series (r ≠ 1) is:

$\displaystyle \large{S_n=\frac{a_1(r^n-1)}{r-1}}$ or $\displaystyle \large{S_n=\frac{a_1(1-r^n)}{1-r}}$

You may use either one of these formulas but I’ll use the first formula.

We are also given that n = 5, meaning we are adding up 5 terms in the series, substitute n = 5 in along with r = 2 and first term = 1.

$\displaystyle \large{S_5=\frac{1(2^5-1)}{2-1}}\\\displaystyle \large{S_5=\frac{2^5-1}{1}}\\\displaystyle \large{S_5=2^5-1}\\\displaystyle \large{S_5=32-1}\\\displaystyle \large{S_5=31}$

Therefore, the solution is 31.

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Summary

If the sequence has common ratio then the sequence or series is classified as geometric sequence/series.

Common Ratio can be found by either multiplying terms with common ratio to get the exact next sequence which can be expressed as $\displaystyle \large{a_{n-1} \cdot r = a_n}$ meaning “previous term times ratio = next term” or you can also get the next term to divide with previous term which can be expressed as: