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LiRa [457]
3 years ago
6

(-4+4i)-(-6+9i) What is the answer to this complex number?

Mathematics
1 answer:
Tanya [424]3 years ago
6 0
The answer is 2-5i. Hope this helps

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How many different triangles, if any, can be drawn with side lengths of 2 cm, 4 cm, and 7 cm?
sveticcg [70]

Answer:                                                                                                                         None of them.

5 0
3 years ago
Read 2 more answers
Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you
Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}

With \pi, we should expect \cos x. If \tan x=\dfrac8{15}, then

\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}

Also,

\pi

so we should expect \tan\dfrac x2 and

\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4

You seem to be taking

\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}

but this is not an identity.

###

8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

\dfrac{\sqrt6-\sqrt2}4

###

9. Use the identity from (7). \dfrac{7\pi}{12} lies in the second quadrant, so

\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3

###

12.

\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}

=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}

=\dfrac{\cos x}{\sin x}

=\dfrac{\csc x}{\sec x}

###

13.

\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}

=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}

=\dfrac1{\cos x}

=\sec x

###

14.

\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)

=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}

=2

###

15.

\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}

=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}

=\cos2\theta

3 0
3 years ago
The problem is 14?minus 85
denpristay [2]

Answer:

-71, and also 71

Step-by-step explanation:

It's Simple Subtraction or if your doing 85-14 its gonna be 71

3 0
4 years ago
What is a true statement about the number 1?
AysviL [449]

Answer:

It is a whole number

Step-by-step explanation:

It has not decimals or fractions

6 0
3 years ago
Read 2 more answers
If a +b+c=9 and a square +b square +square =1 then ab+bc+ca
MatroZZZ [7]

Answer:

ab + bc + ca = 40

Step-by-step explanation:

Using the expansion identity

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) , that is

9² = 1 + 2(ab + bc + ca)

81 = 1 + 2(ab + bc + ca) ← subtract 1 from both sides

80 = 2(ab + bc + ca) ← divide both sides by 2

40 = ab + bc + ca

4 0
3 years ago
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