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2 years ago

15

Solve: k/3+4-5k=-2k (LOOK AT PICTURE)

1 answer:

stepan [7]2 years ago

3 0

The answer for this is k=3/2

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Trig: Which complex number's graph is shown?

Andrei [34K]

ANSWER

$2 \sqrt{2} ( \cos( \frac{7\pi}{4} + i \sin( \frac{7\pi}{4} ) )$

EXPLANATION

The complex number shown has coordinates (2,-2)

or

$z = 2 - 2i$

The modulus is

$|z| = \sqrt{ {2}^{2} + {( - 2)}^{2} }$

$|z| = \sqrt{ 8} = 2 \sqrt{2}$

The argument is

$\theta= \tan^{ - 1} ( \frac{ - 2}{2} )$

$\theta= \tan^{ - 1} ( -1) = \frac{7\pi}{4}$

The polar form is

$2 \sqrt{2} ( \cos( \frac{7\pi}{4} + i \sin( \frac{7\pi}{4} ) )$

The first option is correct.

3 0

3 years ago

David rented a truck for one day. There was a base fee of $16.95 , and there was an additional charge of 98 cents for each mile

Nadusha1986 [10]

136.51=16.95+.98x

119.56=.98x

x=122

3 0

2 years ago

Please help me with the whole problem.

lions [1.4K]

Answer:

1. $155 ≥ $31.50p + $13.25

2. p ≤ 4

6 0

2 years ago

50+30 x12 20 POINTS GUYS YESSSSSSSSSSSSSSSSSSSSSSS

daser333 [38]

Answer:

960

Step-by-step explanation:

50+30= 80

80x12=960

8 0

3 years ago

Read 2 more answers

[tex]cos {}^{4} α+sin {}^{4} α= \frac{1}{4} (3+cos4α)<br>Prove:<br>

asambeis [7]

Given:

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

To prove:

The given statement.

Proof:

We have,

$\cos^4 \alpha+\sin^4\alpha=\dfrac{1}{4}(3+\cos 4 \alpha)$

$LHS=\cos^4 \alpha+\sin^4\alpha$

$LHS=(\cos^2 \alpha)^2+(\sin^2 \alpha)^2$

$LHS=(\cos^2 \alpha+\sin^2\alpha)^2-2\sin ^2\alpha\cos^2 \alpha$ $[\because a^2+b^2=(a+b)^2-2ab]$

$LHS=(1)^2-2(1-\cos^2 \alpha)\cos^2 \alpha$ $[\because \cos^2 \alpha+\sin^2\alpha=1]$

$LHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

Now,

$RHS=\dfrac{1}{4}(3+\cos 4 \alpha)$

$RHS=\dfrac{1}{4}[3+(2\cos^2 2\alpha-1)]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2\cos^2 2\alpha]$

$RHS=\dfrac{1}{4}[2+2(2\cos^2 \alpha-1)^2]$ $[\because \cos 2\theta=2\cos^2\theta -1]$

$RHS=\dfrac{1}{4}[2+2(4\cos^4 \alpha-4\cos \alpha+1)]$ $[\because (a-b)^2=a^2-2ab+b^2]$

$RHS=\dfrac{1}{4}[2+8\cos^4 \alpha-8\cos \alpha+2]$

$RHS=\dfrac{1}{4}[4+8\cos^4 \alpha-8\cos \alpha]$

$RHS=1+2\cos^4 \alpha-2\cos \alpha$

$RHS=1-2\cos^2 \alpha+2\cos^4 \alpha$

$LHS=RHS$

Hence proved.

8 0

3 years ago