Arrange the four expressions in ascending order of their values when x = -2.

Answers

Answer 1

Answer:

Answer:

3-1 2-4 1-3 4-2

Step-by-step explanation:

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At swim practice, several swimmers take turns, swimming 100 meters each. The total distance they swim is less than 1,200 meters. Which inequality can be solved to determine, s, the possible number of swimmers?1,200 greater-than 100 s
1,200 less-than 100 s
1,200 s greater-than 100
1,200 s less-than 100

Answers

Answer:

1,200 greater-than 100 s

Step-by-step explanation:

Each swimmer swims 100m.

So the total distance they swim is given by:

T = 100s.

The total distance they swim is less than 1,200 meters.

This means that:

T < 1200

100s < 1200

1200 greater than 100s.

So the correct answer is:

1,200 greater-than 100 s

Answer:

Step-by-step explanation:

The students at Midtown Middle school sold flowers as a fundraiser in September and October. In October, they charged $1.50 for each flower. The October price was a 20% increase of the September price. Part A: What was the price of the flowers in September? Part B: The seventh-grade class earned 40% of the selling price of each flower. In September, they sold 900 flowers. In October, they sold 700 flowers. Did they earn more money in September or October? How much more?

Answers

Let the price of flowers in September be x, then (100 + 20)/100 * x = $1.50
120/100 * x = $1.50
1.2x = $1.50
x = $1.50/1.2 = $1.25
The price of flowers in September is $1.25

Total money realised in September = $1.25 x 900 = $1,125
40% earned = 0.4 x $1,125 = $450

Total money realised in October = $1.50 x 700 = $1,050
40% earned = 0.4 x $1,050 = $420

The class earned more money in September and they earned $30 more than in October.

Solve the given system using the method of Example 3.25. x1 − x2 = 1 3x1 + x2 = 3 x = 1 0 Correct: Your answer is correct.

Answers

Answer:

$x_1 = 1$

$x_2=0$

Step-by-step explanation:

The question is incomplete as the method is not given.

However, the question can still be solved.

Given

$x_1 - x_2 = 1$

$3x_1 + x_2 = 3$

Make xi the subject in the first equation

$x_1 = 1 + x_2$

Substitute 1 + x2 for xi in the second equation

$3(1+x_2)+x_2 = 3$

Open bracket

$3+3x_2+x_2=3$

$3+4x_2=3$

Collect Like Terms

$4x_2 = 3-3$

$4x_2 =0$

Solve for x2

$x_2=0/4$

$x_2=0$

Recall that:

$x_1 = 1 + x_2$

$x_1 = 1+0$

$x_1 = 1$

Final answer:

The solution to the given system of equations is obtained through substitution. The process involves replacing a variable in one equation with an expression from the other. The final solutions are x1=1 and x2=0.

Explanation:

The system of equations in question is:

1) x1 - x2 = 1

2) 3x1 + x2 = 3

The method to solve this system is through substitution or elimination. First, rewrite the first equation x1 = x2 + 1. This allows us to substitute x1 - 1 for x2 in the second equation, yielding 3(x2 + 1) + x2 = 3, simplifying to 4x2 + 3 = 3. Solving for x2, we get x2 = 0. Substituting x2 into x1 = x2 + 1, we conclude that x1 = 1. Thus, the solution to the system is x1 = 1, x2 = 0.

Learn more about System of Equations here:

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Find the domain of f and f −1 and its domain. f(x) = ln(ex − 3). (a) Find the domain of f. (Enter your answer using interval notation.) (−2,[infinity]) (b) Find f −1. f −1(x) = x+ln(3)

Answers

Answer:

a.Domain of f=(1.099, $\infty)$

b. $f^(-1)(x)=ln(e^x+3)$

Step-by-step explanation:

Let $y=f(x)=ln(e^x-3)$

We know that domain of ln x is greater than zero

$e^x-3>0$

Adding 3 on both sides of inequality

$e^x-3+3>0+3$

$e^x>3$

Taking on both sides of inequality

$lne^x>ln 3$

$x>ln 3$ =1.099

By using $lne^x=x$

Domain of f=(1.099, $\infty)$

Let $y=f^(-1)(x)=ln(e^x-3)$

$e^y=e^x-3$

By using property $lnx=y\implies x=e^y$

$e^x=e^y+3$

Taking ln on both sides of equality '

$lne^x=ln(e^y+3)$

$x=ln(e^y+3)$

Replace x by y and y by x

$y=ln(e^x+3)$

Substitute y= $f^(-1)(x)$

$f^(-1)(x)=ln(e^x+3)$

Find the arc length of the curve below on the given interval. y equals one third (x squared plus 2 )Superscript 3 divided by 2y= 1 3x2+23/2 on [00,66]