Ask question

Ask question

All categories

3 years ago

9

A computer was originally purchased for $900 and undergoes a 14% annual depreciation rate. The computer is now worth $364. The n

umber of years that have passed, t, since the computer was purchased can be found using the following equation:

2 answers:

nekit [7.7K]3 years ago

7 0

Answer:

6 years

Step-by-step explanation:

on edge 2020

nexus9112 [7]3 years ago

5 0

364=900(1-0.14)t is the equation.

You might be interested in

Write an equation in slope-intercept form of the line that passes through the givin point and is parcel to the graph of the give

Serhud [2]

The answer is X=4.

It is both parallel to x=-3 and passes through (4,2)

5 0

4 years ago

A wall in Marcus's bedroom is 8 2/5 feet high and 18 1/3 feet long. If he paints 1/2 of the wall blue, how many square feet will

goldenfox [79]

$\bf \textit{so the area of the wall is then }8\frac{2}{5}\times 18\frac{1}{3}
\\\\\\
\textit{how much is }\frac{1}{2}\textit{ of that product?}\quad \cfrac{8\frac{2}{5}\times 18\frac{1}{3}}{2}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
8\frac{2}{5}\implies \cfrac{8\cdot 5+2}{5}\implies &\cfrac{42}{5}\\\\
18\frac{1}{3}\implies \cfrac{18\cdot 3+1}{3}\implies &\cfrac{55}{3}
\end{cases}
\\\\\\
\cfrac{8\frac{2}{5}\times 18\frac{1}{3}}{2}\implies \cfrac{\frac{42}{5}\cdot \frac{55}{3}}{\frac{2}{1}}\implies \cfrac{\frac{42\cdot 55}{5\cdot 3}}{\frac{2}{1}}\implies \cfrac{42\cdot 55}{5\cdot 3}\cdot \cfrac{1}{2}\implies \cfrac{42\cdot 55\cdot 1}{5\cdot 3\cdot 2}
\\\\\\
\cfrac{2310}{30}\implies 77$

7 0

3 years ago

A company sells cases of Mountain Dew. The revenue generated by selling x cases is described as LaTeX: \text{R(x)=50x} . The cos

Artist 52 [7]

Answer:the person below is correct

Step-by-step explanation:

8 0

3 years ago

Convert the mixed number to an improper fraction. enter your answer below using the slash mark ( / ) as the fraction bar

eduard

The Answer Fam Is 99/10

5 0

4 years ago

Consider the following two ordered bases of R3:

grigory [225]

Answer:

Let $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors $b_1, ..., b_n$ at base A.

The, we case correspond to find the coordinates of vectors of C,

$\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] \}$

at base B.

1. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

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

Using rows operation we obtain the echelon form of the matrix

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

now we use backward substitution

$c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\2\\1\end{array}\right]$

2. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

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

Using rows operation we obtain the echelon form of the matrix

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

now we use backward substitution $c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\3\\2\end{array}\right]$

3. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

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

Using rows operation we obtain the echelon form of the matrix

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

now we use backward substitution $c=-2\\-b+c=2,\; b=-4\\a-2b+2c=2,\; a=-2$

Then the coordinate vector of $\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]$ is $\left[\begin{array}{ccc}-2\\-4\\-2\end{array}\right]$

Then the change of basis matrix from B to C is

$\left[\begin{array}{ccc}4&4&-2\\2&3&-4\\1&2&-2\end{array}\right]$

4 0

4 years ago