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3 years ago

How do you subtract o.8 from 1.5

2 answers:

5 0

Do 1.5-0.8. Go from left to right by doing .5-.8. That is a negative number so borrow from the ones place. 15-8=7. Drop your decimal and you get 0.7.

Iteru [2.4K]3 years ago

3 0

All that is done is that 1.5 is larger than 0.8 and you subtract the smaller decimal from the larger decimal

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How do you find the volume of a circle?

Kazeer [188]

To have a volume, a shape must be 3D, so a circle would probably be a sphere, whose volume is found by $v = \frac{4}{3} \pi r^{3}$ .
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4 years ago

Solve the simultaneous equations by substitution<br> 8y−3x=23<br> 3x=2y+1

Oxana [17]

Answer:

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Step-by-step explanation:

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3 years ago

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

Simplify each expression as much as possible, and rationalize denominators when applicable. √(17/25)=?

amm1812

Answer:

The answer to your question is: $\frac{\sqrt{17}}{5}$

Step-by-step explanation:

$\sqrt{\frac{17}{25} } = \frac{\sqrt{17} }{\sqrt{25}} = \frac{\sqrt{17}}{5}$

7 0

4 years ago

Triangle J K L is shown. Angle J K L is 120 degrees and angle K L J is 40 degrees. The length of K L is 2 and the length of J L

Veseljchak [2.6K]

Answer:

$k \approx 5$ units$

Step-by-step explanation:

In Triangle JKL

$\angle K=120^\circ\\\angle L=40^\circ\\KL=2\\JL=k$

We want to determine the approximate value of k using the law of sines.

$\angle J+\angle K+\angle L=180^\circ $ (Sum of angles in a \triangle)\\\angle J+120^\circ+40^\circ=180^\circ \\\angle J=180^\circ-(120^\circ+40^\circ)=20^\circ$

Using Law of Sines

$\dfrac{k}{\sin K} =\dfrac{j}{\sin J} \\\dfrac{k}{\sin 120} =\dfrac{2}{\sin 20} \\k=\sin 120 \times \dfrac{2}{\sin 20}\\k=5.06\\k \approx 5$ units$

7 0

3 years ago

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