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4 years ago

Ok so can someone pls help me on this. It’s math, and pls explain.

Mathematics

1 answer:

Anika [276]4 years ago

3 0

F.v*king ki11 yourself, you gay piece of $.h1t. Gay b17ches like you burn in hell

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I forgot how to do this ? I need help in interest please help

salantis [7]

I=prt so you multiply principal rate and time to get the answer

6 0

3 years ago

PLEASE HELP BEST ANSWER GETS BRAINLIEST AND 10 POINTS

lubasha [3.4K]

Answer:

94 Minutes.

Step-by-step explanation:Subtract 220 from 314 and you get 94. To check you add 220 to 94 and you get 314. Hope this helped.

7 0

3 years ago

#6: Tell whether the function below is 1 point linear or exponential. * f(x) = -1/4x + 5

Pachacha [2.7K]

Answer:

Linear function

Step-by-step explanation:

Given

$f(x) = -\frac{1}{4}x + 5$

Required

Linear or Exponential

A linear equation has the form: $f(x) = mx + c$ ;

An exponential has the form: $f(x) = ab^x$

By comparing both functions to: $f(x) = -\frac{1}{4}x + 5$

The linear function is similar to $f(x) = -\frac{1}{4}x + 5$

Hence, the given function is a linear function.

8 0

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

How many more is the 3 in 43021.4 then the 3 in 9760

svet-max [94.6K]

this answer is not possible there is no 3 in 9760

3 0

3 years ago