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3 years ago

Mr. Bell earns a yearly salary of $34,567. Mr. Lee earns $22,999 per year. How much more does Mr. Bell earn

Mathematics

1 answer:

bogdanovich [222]3 years ago

7 0

Answer:

34,567-22,999=11,568

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Maggie has 7 tiles with pictures of plants and 2 tiles with pictures of animals. Maggie keeps all the tiles on a mat with the pi

tatuchka [14]

Answer:6/8 - - - - simplified 3/4

Step-by-step explanation: removing the tile she just flipped over would be 6 plants and 2 animals so there would be 8 total tiles and you would putt the remaining tiles over it like so 6/8 meaning out of the remaining 8 tiles 6 of them are plants giving her a 6 in 8 chance or a 3 in 4 chance

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3 years ago

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Choose the correct definition of (n/k) from below

Citrus2011 [14]

Answer:C. $\frac{n!}{k!(n-k)!}$

Step-by-step explanation:

We know that a combination is a collection of the items where the order doesn't matter. It is a way to calculate the total outcomes of an event where order of the outcomes does not matter. It is denoted by C(n,k) or (n,k) or (n/k).

The formula for the number of combinations of n things taken r at a time is given by :-

$C(n,k)=(n/k)=\frac{n!}{k!(n-k)!}$

Hence, C is the right option.

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3 years ago

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Math question! <img src="https://tex.z-dn.net/?f=%20%5Cint%5C%20%7B%20%5Csqrt%7Btan%28x%29%7D%20%7D%20%5C%2C%20dx%20" id

stira [4]

Solution: $\int\limits \: \sqrt{tan(x)} \: dx$
1) pass: make the replacement
 $y = \sqrt{tan(x)} = tan(x)^ \frac{1}{2}$
$dy = \frac{1}{2} *(tan\:x)^{- \frac{1}{2}} * sec^2x\:dx$
$2dy = \frac{1}{y}*(tan^2*x+1)dx$
$2ydy = (y^4+1)dx$
$dx = \frac{2y}{y^4+1} dy$

2) pass: We substitute in the integral of the statement
 $I = \int\limits \sqrt{tanx} \: dx$
$I = \int\limits \:y* \frac{2y}{y^4+1} dy$
$I = \int\limits\: \frac{2y^2}{y^4+1} dy$

3) pass: Using the Gauss Lemma, we will factor the polynomial (y⁴ + 1) knowing that there are no real roots, so we will directly try to factorize into two polynomials of degree 2:

$y^4+1 = (y^2+ay+1)(y^2+cy+1)$
$y^4+1 = y^4+(a+c)y^3 + (ac+2)y^2+(a+c)y+1$
Make the system linear, to find the values of "a" and "c".
 $\left \{ {{a+c=0} \atop {ac+2=0}} \right.$
$a+c = 0 \:\to c=-a$
$ac+2=0\to ac = -2\to a*(-a) = -2\to -a^2 = -2\to a =\sqrt{2}$
$a+c = 0 \:\to c=-a\to c = - \sqrt{2}$
We have:
$(y^4+1)= (y^2+ \sqrt{2y} +1)(y^2- \sqrt{2y} +1)$

4) pass: We will use the partial fractions method, using the fraction from within the integral:
 $\frac{2y^2}{y^4+1} = \frac{Ay+B}{y^2+ \sqrt{2y}+1 } + \frac{Cy+D}{y^2- \sqrt{2y} +1}$
$= \frac{(A+C)y^3+(- \sqrt{2}A+B+ \sqrt{2}C+D)y^2+(A- \sqrt{2}B+C+ \sqrt{2}D)y+(B+D) }{y^4+1}$

$\longrightarrow \left \{ {{A+C=0\to A=-C} \atop { -\sqrt{2}(A-C)+(B+D)=2 }} \right.$
$-\sqrt{2}(A-C)+(B+D)=2 \to - \sqrt{2} *2A+0=2\to A = - \frac{1}{ \sqrt{2} }$
$A+C=0\to C=-A\to C = - (- \frac{1}{ \sqrt{2} } )\to C = \frac{1}{ \sqrt{2} }$
$(A+C)+ \sqrt{2} (D-B)=0\to 0+ \sqrt{2} (D-B)=0 \to B=D=0$
$B+D=0$

5)pass: Adopt what was used above:

$I = \int\limits \frac{ -\frac{1}{ \sqrt{2} }y }{y^2+ \sqrt{2}y+1 } dy+ \int\limits \frac{ \frac{1}{ \sqrt{2} }y }{y^2- \sqrt{2}y+1 } dy$

$I = - \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy }_{I_1}+ \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy }_{I_2}$

6) pass: Now, solve it separately

$I_1 = \int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy$
$2I_1 = \int\limits\frac{2y}{y^2+ \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy$
$2I_1 = \int\limits\frac{2y+ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy$
$2I_1 = ln|y^2+ \sqrt{2}y+1| - \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}+1)^2+1 }dy$
$\boxed{I_1 = \frac{1}{2} ln|y^2+ \sqrt{2} y+1| - \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}+1 )}$

and

$I_2 = \int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy$
$2I_2 = \int\limits\frac{2y}{y^2- \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy$
$2I_2 = \int\limits\frac{2y- \sqrt{2} }{y^2- \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2- \sqrt{2y}+1 }dy$
$2I_2 = ln|y^2- \sqrt{2}y+1| + \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}-1)^2+1 }dy$
$\boxed{I_2 = \frac{1}{2} ln|y^2- \sqrt{2} y+1| + \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}-1 )}$

7) pass: Let's use the expression of $I$

$I = - \frac{I_1}{ \sqrt{2} } + \frac{I_2}{ \sqrt{2} }$

$I = - \frac{ \frac{1}{2}ln|y^2+ \sqrt{2}y+1|- \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} + 1) }{ \sqrt{2} } + \frac{ \frac{1}{2}ln|y^2- \sqrt{2}y+1|+ \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} - 1) }{ \sqrt{2} }$
$I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+$
$+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)$

8) pass: Now, returning to the expression as a function of x, we finally arrive at the final answer:

$I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+$
$+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)$
$I = - \frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2+ \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} +1$
$+\frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2- \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} -1$

Answer:

$\boxed{I = - \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x+\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} +1) +}$
$\boxed{+ \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x-\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} -1) +C}}$ $\end{array}}\qquad\quad\checkmark$

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4 years ago

What is cos(npi/2) and sin(npi/2) in fourier series cos(npi) is (-1)^n and sin(npi)=0 ?

Nat2105 [25]

The problem ask to fourier series of the trigonometric function base on the data you have been given in the problem, so the answer would be cos(pi/2n) = {(-1)^n/2 - if n is even, 0 - if n is odd} and sin(pi/2n) = {0- if n is even, (-1)^(n-1)/2 if n is odd}. I hope you are satisfied with my answer

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4 years ago

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Lawrence performs a survey to determine the number of minutes of exercise seventh-grade athletes get in one week. He gets a list

Delicious77 [7]

He can choose his sample from every third person on each list rather than 10 people.

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3 years ago

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