Find (f-g)(x). What is the domain of f/g
f(x)=2x-5 g(x)=2-x

Answers

Answer 1
Answer:
f(x)=2x-5 , \ \ \ \ g(x)=2-x \n\n(f-g)(x)=f(x)-g(x) = 2x-5 -(2-x)=2x-5-2+x = 3x+5 \n\n\n domain : (f)/(g)=(2x-5)/(2-x)\n\n2-x\neq 0 \n \nx\neq 2 \n \nD=R\setminus \left \{ 2 \right \}

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What best describes how to evaluate a variable expression?

Answers

 To replace each letter with its value, and then finish it by doing the order of the operation/ solve it.

Barry wants to take out a loan from his bank for $1000 to buy bicycle the interest rate on the loan from the bank is 9% he wants to pay the toll amount in three monthly payments what is the amount that vary with pay in interest rate loan

Answers

The required interestpaid by Barry is given as, $22.5.

Given that,
Barry wants to take out a loan from his bank for $1000 to buy a bicycle the interest rate on the loan from the bank is 9% he wants to pay the toll amount in three monthly payments what is the amount that varies with pay in interest rate loan is to be determined.

What is interest?

Simpleinterest is defined as the percentage of earnings on the lending for a period of time

Here,
Principal = 1000
Rate = 9%
Time = 3 / 12 = 0.25 years
So,
Interest = Principal × rate × time
interest = 1000 × 0.09 × 0.25
Interest = $22.5

Thus, the required interestpaid by Barry is given as, $22.5.

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Answer:

22.50

Step-by-step explanation:

formula I=prt

I=1000x0.09x.25

I=22.50

Mr. Mudd gives each of his children $2000 to invest as part of a friendly family competition. The competition will last 10 years. The rules of the competition are simple. Each child can split up his or her $2000 into as many separate investments as they please. The children are encouraged to do their research on types of investments. The initial investments made may not be changed at any point during the 10 years; no money may be added and no money may be moved. Whichever child has made the most money after 10 years will be awarded an additional $10,000. Child Performance of investments over the course of the competition Albert $1000 earned 1.2% annual interest compounded monthly $500 lost 2% over the course of the 10 years $500 grew compounded continuously at rate of 0.8% annually Marie $1500 earned 1.4% annual interest compounded quarterly $500 gained 4% over the course of 10 years Hans $2000 grew compounded continuously at rate of 0.9% annually Max $1000 decreased in value exponentially at a rate of 0.5% annually $1000 earned 1.8% annual interest compounded biannually (twice a year) 1. What is the balance of Albert’s $2000 after 10 years? 2. What is the balance of Marie’s $2000 after 10 years? 3. What is the balance of Hans’ $2000 after 10 years? 4. What is the balance of Max’s $2000 after 10 years? 5. Who is $10,000 richer at the end of the competition?

Answers

The balance of Albert is $2159.07; the balance of Marie is $2244.99, the balance of Hans is $2188.35, and the balance of Max is $2147.40. Marie is $10,000 richer at the end of the competition.

What is Compound interest?

Compound interest is defined as interest paid on the original principal and the interest earned on the interest of the principal.

To determine the balance of Albert’s $2000 after 10 years :

If the amount of $1000 at 1.2 % compounded monthly,

A = P(1 +r/n)ⁿ n = 10 years

here P = $1000 and r = 1.2

A = 1000(1 + 0.001)¹²⁰

A = $1127.43

If Albert $500 losing 2%

So 0.98 × 500 = $490

If $500 compounded continuously at 0.8%

So A = Pe^(rt)

A = 500e^(0.008* 10)

A = 541.6

So the balance of Albert’s $2000 after 10 years :

Total balance = 1127.43 + 490.00+ 541.64 = $2159.07

To determine the balance of Marie’s $2000 after 10 years:

If 1500 at 1.4 % compounded quarterly,

A = 1500(1 + 0.0035)⁴⁰ = $1724.99

If $500 Marie’s gaining 4 %

So 1.04 × 500 = $520.00

So the balance of Marie’s $2000 after 10 years

Total balance = 1724.99 + 520.00 = $2244.99

To determine the balance of Hans’ $2000 after 10 years:

If $2000 compounded continuously at 0.9%

So A = 2000e^(0.009* 10)

A = $2188.3

To determine the balance of Max’s $2000 after 10 years :

If $1000 decreasing exponentially at 0.5 % annually

So A = 1000(1 - 0.005)¹⁰= $951.11

If $1000 at 1.8 % compounded bi-annually

So A = 1000(1 + 0.009)²⁰ = $1196.29

So the balance of Max’s $2000 after 10 years

Total balance = 951.11 + 1196.29 = $2147.40

Therefore, Marie is $10,000 richer at the end of the competition.

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Answer:

Step-by-step explanation:

Albert:

$1000 earned 1.2% annual interest compounded monthly

= 1000 (1+.001)120

(periodic interest = .012/12 ,n is periods = 10yr x 12 mos)

$500 lost 2% over the course of the 10 years

= 500 (.98)

$500 grew compounded continuously at rate of 0.8% annually

= 500 e^008(10) 10 years interest .008 (in decimal form)

Add these three to see how Albert did with his investments

You have six trophies to arrange on the top shelf of the bookcase how many ways are there to arrange the trophies

Answers

Answer:

So, we can do it in 720 ways.

Step-by-step explanation:

We have been given total of six trophies to arrange in the top shelf of the bookcase  

If we need to arrange n things in different ways we can do it in n! ways

Similarly, Here we will do it in 6! ways

n!=n(n-1)(n-2).....1

So, 6!=6 x 5 x 4 x 3 x 2 x 1=720

$6.50 is what percent of $130

Answers

   
\displaystyle \n p = (6.5)/(130) * 100 = (6.5* 100)/(130)= (650)/(130)= \boxed{ 5\% }



Solve for d. k = 5(d - 6)

Answers

5(d-6)=k\ \ \ |use\ distributive\ property:a(b-c)=ab-ac\n\n5d-30=k\ \ \ |add\ 30\ to\ both\ sides\n\n5d=k+30\ \ \ \ |divide\ both\ sides\ by\ 5\n\n\boxed{d=(k+30)/(5)}
k = 5(d - 6)  \text{Equation} 

(k)/(5)=d-6  \text{Divide both sides by 5} 

(k)/(5)+6=d  \text{Add 6 to both sides} 

d= (k)/(5)+6 

d= (k+30)/(5)  \text{Multiply the denominator by the whole number} 

\fbox{Hence Solved}