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3 years ago

-6x^2-5x-5= −6x 2 −5x−5= \,\,-4x^2 −4x 2

Mathematics

1 answer:

aev [14]3 years ago

3 0

Answer:

what is the question? it's not clear

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Use demoivre's theorem to show that;<br>cos4¤ =8 cos4¤ -8cos2¤+1

ss7ja [257]

Answer:

$\cos4( \theta)=Re(\cos4( \theta)+i\sin4( \theta)) \\=Re(\cos ( \theta)+i\sin ( \theta))^4 \\ =Re(cos^4( \theta)+4cos^3( \theta)isin( \theta)+6cos^2( \theta)i^2sin^2( \theta)+4cosi^3( \theta)sin^3( \theta)+i^4sin^4( \theta)) \\cos( 4\theta) =cos^4( \theta)-6cos^2( \theta)sin^2( \theta)+sin^4( \theta) \\ =(1-\sin^2( \theta))^2-6(1-\sin^2( \theta))\sin^2( \theta)+sin^4( \theta) \\ = 1-2\sin^2( \theta)+\sin^4( \theta)-6\sin^2( \theta)+6\sin^4( \theta)+\sin^4( \theta) \\ =8\sin^4( \theta)-8\sin^2( \theta)+1$

7 0

3 years ago

Which of the following equations passes through the point (0, -6) and is parallel to

LiRa [457]

Step-by-step explanation:

Y = 3 X-6

... . .... ... ...

3 0

3 years ago

50 points!!! please have serious answers

Darina [25.2K]

Median: Put the list in order from least to greatest.

$20, $20, $23.50, $25, $26.50

$20, $23.50, $25

$23.50 is the median. I took one from each side until I had one left.

Mean: Get the sum of them all and then divide by the number of earnings ( in this case )

$25 + $26.50 = $51.50 $51.50 + $20 = $71.50 $71.50 + $23.50 = $95

$95 + $20 = $115 $115/5 = $23

$23 is the mean.

Mode: Is the one that finds the frequency or in lamen terms it is the number which appears most often.

There's no steps for finding the expect counting how many times that number appeared.

$20 is the mode.

4 0

3 years ago

Is 83 ft larger than 786 in???

prohojiy [21]

83 ft is larger than 786 in.

To convert feet to inches, multiply 83ft by 12 because there are 12 inches in a foot.

83 x 12 = 996

996 in > 786 in

7 0

3 years ago

Read 2 more answers

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

4 years ago