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3 years ago

For better clarification I guess

Mathematics

1 answer:

marissa [1.9K]3 years ago

6 0

Answer:

1.) $=\frac{\sqrt{3}}{3}$

2.) $42\sqrt{30}$

Step-by-step explanation:

1.) $\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{3}}{\sqrt{3}}$

2.) $\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}\sqrt{b}=\sqrt{a\cdot b}$

$\sqrt{6}\sqrt{5}=\sqrt{6\cdot \:5}$

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Find the imaginary part of\[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]

iren [92.7K]

Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

$x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6$

First, we can expand this power using the binomial theorem:

$(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}$

After that, we can apply De Moivre's theorem to expand each summand: $(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)$

The final step is to find the common factor of i in the last expansion. Now:

$x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6$

$=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6$

$=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))$

The last part is to multiply these factors and extract the imaginary part. This computation gives:

$Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288$

$Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288$

(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)

A calculator simplifies the imaginary part Im(x⁶) to 0

4 0

3 years ago

suppose that a small pump can empty a swimming pool in 60 hours and that a large pump can empty a pool in 40 hours. working toge

Aloiza [94]

I would assume that you would subtract 40 from 60 and get 20. (60-40=20) So I'd say 20 hours.

8 0

3 years ago

What is the equation of a circle whose center is at the origin and whose radius is 9?

Rashid [163]

A=πr2=π·92≈254.469. I think this is what your looking for

3 0

4 years ago

Read 2 more answers

An isosceles triangle has an angle that measures 94º. Which other angles could be in that

Nataly [62]

Answer:

43°

Step-by-step explanation:

In a isosceles triangle, two angles HAVE to be equal.

This means that 94 can be one of the angles that are equal, or the angle that is not equal to another angle.

Let's start with the fact that 94 can be one of the angles that are equal. (Note: angles in a triangle add up to 180)

94+94+x=180

188+x=180

There is NOT a possible value for x in this situation, as you cannot have a negative value of an angle. Therefore we must try the other situation where 94 is the angle that is not equal to another angle.

x+x+94=180

2x+94=180

Subtract 94 from both sides

2x=86

Divide both sides by 2

x=43

Therefore the answer is 43°

4 0

2 years ago

The sum (5z+4) +(3z-6)

Rzqust [24]

Answer:

(5z + 4) + (3z - 6) = 8z - 2

Step-by-step explanation:

(5z + 4) + (3z - 6)

Combine them by their like terms.

(5z + 3z) + (4 + (-6))

Add 5z and 3z.

8z + (4 + (-6))

Add 4 and -6.

8z + (-2)

Simplify.

8z - 2

6 0

3 years ago