Find solution to logarithm.

Answers

Answer 1

Answer: $\log5x+\log(x-1)=2\n D:5x>0 \wedge x-1>0\n D:x>0 \wedge x>1\n D:x>1\n \log5x(x-1)=2\n 10^2=5x^2-5x\n 5x^2-5x-100=0\n x^2-x-20=0\n x^2-5x+4x-20=0\n x(x-5)+4(x-5)=0\n (x+4)(x-5)=0\n x=-4 \vee x=5\n -4\not \in D\Rightarrow x=5$

Answer 2

Answer: $log_pa=b\ \ \ \Leftrightarrow\ \ \ p^b=a\n\nloga+logb=log(a\cdot b)\ \ \ \Rightarrow\ \ \ D:\ a>0\ \ \ and\ \ \ b>0\n ----------------------------- \nlog(5x)+log(x-1)=2\ \ \ \Rightarrow\ \ \ D:\ 5x>0\ \ \ and\ \ \ x-1>0\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x>0\ \ \ and\ \ \ \ \ \ \ \ x>1\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ D=(1;+\infty)\n\nlog[5x\cdot(x-1)]=2\ \ \ \Leftrightarrow\ \ \ 5x(x-1)=10^2$

$5x^2-5x=100\ /:5\n\nx^2-x-20=0\n\nx^2-5x+4x-20=0\n\nx(x-5)+4(x-5)=0\n\n(x-5)(x+4)=0\ \ \ \Leftrightarrow\ \ \ x-5=0\ \ \ or\ \ \ x+4=0\n.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=5\ \ \ or\ \ \ \ \ \ \ \ x=-4\n\n5\in D;\ \ \ -4\notin D\n\nAns.\ x=5$

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Steve invests $1,800 in an account that earns 3.7% annual interest, compounded continuously. What is the value of the account after 10 years? Round your answer to the nearest dollar. A. about $2,466 B. about $2,589 C. about $2,601 D. about $2,606

Answers

Answer:

B. about $2,589

Step-by-step explanation:

Given data

P=$1,800

R= 3.7%

T=10years

The expression for the compound interest is given as

A= P(1+r)^t

substitute

A= 1800(1+0.037)^10

A= 1800(1.037)^10

A=1800*1.43809495884

A=$2588.570

Hence the answer is B. about $2,589

Explain this algebraic expression in your own words: 3x - y

Answers

Answer:

You have a number multiplied by the variable x, so three times some value x. then you have (-y) taking the value of 3x and subtracting the value of (y) from it. The expression depends on the values of x and y you plug into it.

Step-by-step explanation:

for example, 3(4) - (6) = 12-6 = 6

Please help asap 25 pts

Answers

Your answer is the first one which is the graph is of exponential growth

It is an exponential growth!

Nemecek Brothers make a single product on two separate production lines, A and B. Its labor force is equivalent to 1000 hours per week, and it has $3000 outlay weekly on operating costs. It takes 1 hour and 4 hours to produce a single item on lines A and B, respectively. The cost of producing a single item is $5 on line A and $4 on line B. (a) Write the inequality that expresses the labor information. (b) Write the inequality that expresses the cost information.

Answers

Answer:

(a) The inequality for the number of items, x, produced by the labor, is given as follows;

250 ≤ x ≤ 600

(b) The inequality for the cost, C is $1,000 ≤ C ≤ $3,000

Step-by-step explanation:

The total time available for production = 1000 hours per week

The time it takes to produce an item on line A = 1 hour

The time it takes to produce an item on line B = 4 hour

Therefore, with both lines working simultaneously, the time it takes to produce 5 items = 4 hours

The number of items produced per the weekly labor = 1000/4 × 5 = 1,250 items

The minimum number of items that can be produced is when only line B is working which produces 1 item per 4 hours, with the weekly number of items = 1000/4 × 1 = 250 items

Therefore, the number of items, x, produced per week with the available labor is given as follows;

250 ≤ x ≤ 1250

Which is revised to 250 ≤ x ≤ 600 as shown in the following answer

(b) The cost of producing a single item on line A = $5

The cost of producing a single item on line B = $4

The total available amount for operating cost = $3,000

Therefore, given that we can have either one item each from lines A and B with a total possible item

When the minimum number of possible items is produced by line B, we have;

Cost = 250 × 4 = $1,000

When the maximum number of items possible, 1,250, is produced, whereby we have 250 items produced from line B and 1,000 items produced from line A, the total cost becomes;

Total cost = 250 × 4 + 1000 × 5 = 6,000

Whereby available weekly outlay = $3000, the maximum that can be produced from line A alone is therefore;

$3,000/$5 = 600 items = The maximum number of items that can be produced

The inequality for the cost, C, becomes;

$1,000 ≤ C ≤ $3,000

The time to produce the maximum 600 items on line A alone is given as follows;

1 hour/item × 600 items = 600 hours

The inequality for the number of items, x, produced by the labor, is therefore, given as follows;

250 ≤ x ≤ 600

(a) The inequality for the number of items, x, produced by the labor, is given as follows;

250 ≤ x ≤ 600

(b) The inequality for the cost, C is $1,000 ≤ C ≤ $3,000

What is inequality?

Inequality is a statement shows greater the, greater then equal to, less then,less then equal to between two algebraic expressions.

The total time available for production = 1000 hours per week

The time it takes to produce an item on line A = 1 hour

The time it takes to produce an item on line B = 4 hour

Therefore, with both lines working simultaneously, the time it takes to produce 5 items = 4 hours

The number of items produced per the weekly labor = 1000/4 × 5 = 1,250 items

The minimum number of items that can be produced is when only line B is working which produces 1 item per 4 hours, with the weekly number of items = 1000/4 × 1 = 250 items

Therefore, the number of items, x, produced per week with the available labor is given as follows;

250 ≤ x ≤ 1250

Which is revised to 250 ≤ x ≤ 600 as shown in the following answer

(b) The cost of producing a single item on line A = $5

The cost of producing a single item on line B = $4

The total available amount for operating cost = $3,000

Therefore, given that we can have either one item each from lines A and B with a total possible item

When the minimum number of possible items is produced by line B, we have;

Cost = 250 × 4 = $1,000

When the maximum number of items possible, 1,250, is produced, whereby we have 250 items produced from line B and 1,000 items produced from line A, the total cost becomes;

Total cost = 250 × 4 + 1000 × 5 = 6,000

Whereby available weekly outlay = $3000, the maximum that can be produced from line A alone is therefore;

$3,000/$5 = 600 items = The maximum number of items that can be produced

The inequality for the cost, C, becomes;

$1,000 ≤ C ≤ $3,000

The time to produce the maximum 600 items on line A alone is given as follows;

1 hour/item × 600 items = 600 hours

The inequality for the number of items, x, produced by the labor, is therefore, given as follows;

250 ≤ x ≤ 600

Hence the inequality for the number of items, x, produced by the labor, is 250 ≤ x ≤ 600 and the inequality for the cost, C is $1,000 ≤ C ≤ $3,000

To know more about Inequality follow

brainly.com/question/24372553

Please help!!!!!You just purchased a cellular phone and are trying to decide the best cellular phone company in which to give your business. When you contacted the Talks-A-Lot Company, they were offering a monthly plan of $40 for 600 minutes and $0.35 for each minute exceeding the 600 minutes. In the Sunday paper you see an ad for the Chat-Away Company, which offers a monthly plan of $50 for 600 minutes and $0.10 for each minute exceeding the 600 minutes. How many minutes would you have to talk over and above the 600 minutes for the cost to be the same with both companies and what would be the equal cost?

Answers

Well first of all that is a lot of money at least for me but like yeah And second of all I’m sorry because I’m not sure

What is the difference between -10 and 17 on a number line?

Answers

The difference would be 7

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