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3 years ago

1. Write two different equations whose solutions are 5 .

1 answer:

3 0

Two equations that have the solution 5:
3y = 2y + 5
2y - y = -3 + 8

To solve an equation is to isolate a variable and find its specific value.

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The temperature in Waterloo was − 5°C at sunrise. By noon the temperature had risen by 12 Celsius degrees. What was the temperat

allochka39001 [22]

The answer is 7° Celsius.

6 0

3 years ago

Read 2 more answers

Determine whether the improper integral converges or diverges, and find the value if it converges.

salantis [7]

Answer:

Integral will be diverging in nature

Step-by-step explanation:

We have given integral $\int_{5}^{\infty}\frac{3}{2}x^2dx$

Now after solving the integral

$[\frac{3}{2}\frac{x^3}{3}]$ limit from 5 to infinite

So $[\frac{3}{2}\frac{\infty^3}{3}]-[\frac{3}{2}\times \frac{5^3}{3}]=\infty$

As after solving integral we got infinite value so integral will be diverging in nature

7 0

2 years ago

SHOW YOUR WORK, <br><br>PLEASE HELP!!!!!!!!!

Shkiper50 [21]

To find the inverse, we swap the variables y and x, then solve for the new y.

3a. $y=\frac{3}{x-1}$

Swapping the variables: $x=\frac{3}{y-1}$
Solving for y: $x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x}$
The domain of this inverse is $x ≠ 0$ .
3b. $y=x^2-1$

Swapping: $x = y^2 - 1$
Solving for y: $y^2 = x + 1 \\ y = \sqrt{x+1}$
The domain of this inverse is $x ≥ -1$ .
3c. $y=\sqrt[3]{\frac{x-7}{3}}$
Swapping: $x=\sqrt[3]{\frac{y-7}{3}}$
Solving for y: $x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7$
The domain of this inverse is all real numbers.
4a. $y=\frac{3}{x-1}$ , $y=1+\frac{3}{x}$
$y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x$
$y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x$

4c. $y=\sqrt[3]{\frac{x-7}{3}}$ , $y=3x^3+7$
$y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x$
$y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x$

3 0

3 years ago

A new car is purchased for 18000 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be,

leonid [27]

Answer:

The value of car after 14 years is $ 2,363.04

Step-by-step explanation:

Given as :

The price of new car = N = $ 18,000

The rate of depreciation of the value of the car = R = 13.5 % per year

Let The value of car after 14 years = $ x

The time period = 14 years