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2 years ago

3x+15=3(x+5) help please?

Mathematics

2 answers:

bearhunter [10]2 years ago

8 0

$3x+15=3(x+5)\\\\
3x+15=3x+3*5\\\\
3x+15=3x+15\\\\0=0\\\\LHS=RHS\\\\ There\ are\ infinitely\ many\ solutions.$

UkoKoshka [18]2 years ago

3 0

3x+15=3(x+5)

3x+15=3x+15

3x-3x=15-15

Therefore x can be equal to any value.

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The rent for an apartment was $6,600 per year in 2012. If the rent increased at a rate of 4% each year thereafter, use an expone

yan [13]

Answer:

$8,029.91

Step-by-step explanation:

You have to use the exponential growth equation:

A=P(1+r)^t

Where A is the final amount, P is the initial amount, r is the rate of increase, and t is the time in years. So:

A=6,600(1+0.04)^5

(The rate has to be changed into a decimal)

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2 years ago

You roll a 6 sided cube, what is the probability of rolling a multiple of 2 or a number greater than 3?

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3 years ago

Read 2 more answers

7g + 11 subtracted from 12

BlackZzzverrR [31]

A subtracted from b means b-a

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5 0

2 years ago

This is Matrix for pre calc

gavmur [86]

The given system of equations in augmented matrix form is

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]$

If you need to solve this, first get the matrix in RREF:

Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]$

Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]$

Add 164(row 3) to -91(row 4):

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]$

Multiply row 4 by 1/13080:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]$

Add -153(row 4) to row 3:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]$

Multiply row 3 by -1/91:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add 6(row 3) and -8(row 4) to row 2:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Multiply row 2 by 1/5:

$\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]$

Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

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

Multiply row 1 by 1/3:

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

So the solution to this system is $(w,x,y,z)=(1,-2,4,-3)$ .

6 0

3 years ago