which of the following is the equation of a line ine slope-intercept form for a line with slope=5 and intercept=(0,-3)

damaskus [11]

Answer:

(-infinity, +infinity)

Step-by-step explanation:

Even tho' f(x) = 5x has only one term, it's still classified as a polynomial. All polynomials are defined on all x: (-infinity, +infinity).

8 0

3 years ago

Read 2 more answers

Solve 20X10 to the third power

LiRa [457]

I believe 8,000,000. I'm probably wrong though.

5 0

3 years ago

What is the interest rate if a principal of $572 earns $228.80 in interest in four years?

Marianna [84]

Answer:

the answer is 10% if it is simple interest you're talking about.

8 0

2 years ago

Find the reduced row echelon form of the following matrices and then give the solution to the system that is represented by the

GaryK [48]

Answer:

a)

Reduced Row Echelon:

$\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]$

Solution to the system:

$x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}$

b)

Reduced Row Echelon:

$\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]$

Solution to the system:

$x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}$

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

Step-by-step explanation:

To find the reduced row echelon form of the matrices, let's use the Gaussian-Jordan elimination process, which consists of taking the matrix and performing a series of row operations. For notation, R_i will be the transformed column, and r_i the unchanged one.

a) $\left[\begin{array}{cccc}0&4&7&0\\2&1&0&0\\0&3&1&-4\end{array}\right]$

Step by step operations:

1. Reorder the rows, interchange Row 1 with Row 2, then apply the next operations on the new rows:

$R_1=\frac{1}{2}r_1\\R_2=\frac{1}{4}r_2$

Resulting matrix:

$\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right]$

2. Set the first row to 1

$R_3=-3r_2+r_3$

Resulting matrix:

$\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]$

3. Write the system of equations:

$x_1+\frac{1}{2}x_2=0\\x_2+\frac{7}{4}x_3=0\\x_3=-4$

Now you have the reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

$x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}$

b)

$\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right]$

1. $R_2=-2r_1+r_2\\R_3=-r_1+r_3$

Resulting matrix:

$\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]$

2. Write the system of equations:

$4x_1+3x_2=7\\2x_3=-17$

Now you have the reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

$x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}$

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

7 0

3 years ago

He Mercedes S-Class luxury vehicle depreciates by 32.4% after just one year. The starting price in 2015 was $70,000.

kumpel [21]

the depreciation amount after one year for the Mercedes S-Class vehicle is $22,680 .

<u>Step-by-step explanation:</u>

Here we have , he Mercedes S-Class luxury vehicle depreciates by 32.4% after just one year. The starting price in 2015 was $70,000. We need to find What is the depreciation amount after one year for the Mercedes S-Class vehicle. Let's find out:

The starting price in 2015 was $70,000 , let's find price after one year i.e. in 2016 when Mercedes S-Class luxury vehicle depreciates by 32.4% ,According to question following equation is framed as :

⇒ $Price - \frac{Price(32.4)}{100}$

⇒ $70,000 - \frac{70,000(32.4)}{100}$

⇒ $70,000 - 700(32.4)$

⇒ $70,000 - 22,680$

⇒ $\$47,320$

Therefore , the depreciation amount after one year for the Mercedes S-Class vehicle is $22,680 .

4 0

3 years ago