How do I solve the attached problem step by step? and what are the solution?I am behind in the course and must submit this assignment today and and do not have much time to do it. So i will be very thankfull for help :-)The exercise need to be solve without relying on the existence of a gcd (greatest common divisor) in exercise (i) and (ii).
See explanation
If you have memorized that##log2=0.3##you can follow this way##log5=log(10/2)=1-log2=1-0.3=0.7##
If you want a general way to find logarithms without using calculators or tables, you could use this formula:##(1/2)ln|(1+x)/(1-x)|=f(x)=x+x^3/3+x^5/5+…##And##logy=lny/ln10=2/ln10*(1/2*ln|y|)## => ##logy=0.869*(1/2*ln|y|)## where ##y=(1+x)/(1-x)##(Note1: you can use ##2/ln10= 0.868589## with the precision you like. Using two terms of the series, 0.869 has a proper level of precision. Note 2: the values of x must be smaller than 1.)
We can’t calculate ##log5## directly because##(x+1)/(1-x)=5## => ##x+1=5-5x## => ##6x=4## => ##x=1.5##And the series doesn’t converge when ##x>1##
But since ##5=2*2.5##for ##y_1=2 ->(x+1)/(1-x)=2## => ##x+1=2-2x## => ##x=1/3~=0.3333####f(x=1/3)=1/3+1/3^3*1/3=1/3+1/81=0.3333+0.0123=0.3456##
for ##y_2=2.5 -> (x+1)/(1-x)=2.5## => ##x+1=2.5-2.5x## => ##3.5x=1.5## => ##x=3/7~=0.4286##Of course we can use this ##x=0.4286##. But perhaps there is an easier way (without a calculator we need to think of this) such as:
Considering that ##5=2^2*1.25## (and since we have already calculated ##f(x=1/3)##):for ##y_2=1.25 -> (x+1)/(1-x)=1.25## => ##x+1=1.25-1.25x## => ##2.25x=0.25## => ##x=25/225=1/9~=0.1111####f(x=1/9)=0.1111+1/9^3*1/3=0.1111+1/729*1/3=1/9+1/2187=0.1111+0.0005=0.1116##(as to the number ##0.0005## just remember that ##10/2=5##)
Using the results above##log5=0.869(2*f(x=2)+f(x=1.25))=0.869(2*0.3456+0.1116)=0.869(0.6912+0.1116)=0.869*0.8028=0.6976332## or ##0.698## in 3 decimals
We should be aware that this last estimate is smaller than the correct result.
(In fact ##log5=0.6990##)
How do I select the best answer for you course hero tutors to be paid? I keep selecting most helpful is that what I’m meant to do? I keep getting asked to select the best answer even after selecting most helpful!
equipment 2800
cr a/c payable 3000
04
dr: cash 1400
cr: service revenue 1400
07
dr: land 22000
cr: cash 22,000
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dr: a/c receivable 1000
cr service revenue 1000
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dr: a/ payable 2800
cr: cash 2800
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dr: telephone bill payable 150
cr : cash 150
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22
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cr: cash 170
29
dr: cash 1600
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31
dr: salary payable 1900
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cr: cash 2,000
Common Stock2Bal 3Bal 4000040000 Bal Bal Service Revenue41129 2800Bal20028007BalAccounts Receivable100018 22000 50031Bal 1900Dividend Payable312000 150Bal Water &…
How do I post the following transaction:The custodian of the pety cash fund submits the following receipts for reinbursment and reports a cash-on-hand count of $8.Postage Stamps Used……..$38United Parcel (Freight-out)$23C.O.D. Postage (Freightout)$51CHristmas Decorations……$30
You use it to determine the polynomial function.
We can use it for higher degree polynomials, but let’s use a cubic as an example. Suppose we have the : -3, 2.5, and 4. So:
##x=-3####x+3=0##
##x=2.5####x=5/2####2x=5## multiply both sides by denominator##2x-5=0##
##x=4####x-4=0##
So, the polynomial function is ##P(x)=(x+3)(2x-5)(x-4)##. Note that we can leave the second root as ##(x-2.5)##, because a proper polynomial function has integer coefficients. It’s also a good idea to put this polynomial into standard form:
##P(x)=2x^3-7x^2-19x+60##
The common mistake in this problem is the sign of the roots. So make sure you do the individuals steps to avoid this mistake.
How do I make it to where I can have the user input s1 and s2. I got the following, but I don’t get the same values when I declare my variables this way:
int main(){
const char *s1 =””;
std::string input1 = s1;
const char *s2 =””;
std::string input2 = s2;
std::cout << “Enter s1: “;
std::getline(std::cin,input1);
int length1 = input1.length();
//std::cout << length1;
std::cout << “Enter s2: “;
std::getline(std::cin,input2);
int length2 = input2.length();
int y = LexDiff(s1,s2);
std::cout << y;
return 0;
}
= LexComp(abc,acz) # = () – () = LexComp(abc,abz) # = () – () + = LexComp(abf,abc) # + = () – () = LexComp(abc,abc) = LexComp(ab,abc) # = () – ()+ = LexComp(abc,ab) # + = () – ()+ = LexComp(abc,abD) = LexComp(abc,abd) = LexComp(aBc,abc) # = () – ()
How do I use the gordon growth model for the following question? “A company expects to have a constant growth rate of 5 percent. It recently paid a dividend of $2.00 per share, and its required rate of return on equity is 15 percent. What is the current price of this company’s stock?”
How do I make an SQL query for something with multiple conditions? for example: Write
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(EECS442
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Let ##(2x+3) ^3## be a given binomial.
From the binomial expression, write down the general term. Let this term be the r+1 th term. Now simplify this general term. If this general term is a constant term, then it should not contain the variable x. Let us write the general term of the above binomial.##T_(r+1)## = ##”” ^3 C_r## ##(2x)^(3-r)## ##3^r##
simplifying, we get, ##T_(r+1)##= ##”” ^3 C_r## ##2^(3-r)## ##3^r## ##x^(3-r)##
Now for this term to be the constant term, ##x^(3-r)## should be equal to 1.Therefore, ##x^(3-r)##= ##x^0##=> 3-r =0=> r=3
Thus, the fourth term in the expansion is the constant term. By putting r=3 in the general term, we will get the value of the constant term.
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