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4 years ago

One number is 3 more than 2 times the other, and their sum is 27. Find the numbers.

Mathematics

2 answers:

TEA [102]4 years ago

4 0

Answer:

Option 2 is correct

Step-by-step explanation:

One number is 2 times another number plus 3. Their sum is 21.

"One number is 2 times another number plus 3" translated to

x = smaller number = another number

It is also given that: Their sum is 21.

Combine like terms:

3x+3 = 21

eimsori [14]4 years ago

3 0

Answer:

I do questions like these everyday so I have too much experience. Let me explain step by step for you.

Brainliest?

First lets set 2 variables x and y

Lets make 2 equations.

x=3+2*y

Thats because it says 'x' is 3 more (+) than 2 times(*) 'y'

Now lets set second, we know both of them add up to 27.

x+y = 27

Since we know what x is equal to (look above equation)

We can replace it.

x is replaced with 3+2*y

3+2y+y = 27

3+4y = 27

Simplify 27-3 = 24

24/4 = 6

Now lets plug in for x

3+2*6 = 15

15 - x

6 - y

:))

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3 years ago

Read 2 more answers

What is the determinant of the coefficient matrix of this system? <br> 4x-3y=-8 <br> 8x-3y=12

gladu [14]

The coefficient matrix is build with its rows representing each equation, and its columns representing each variable.

So, you may write the matrix as

$\left[\begin{array}{cc}\text{x-coefficient, 1st equation}&\text{y-coefficient, 1st equation}\\\text{x-coefficient, 2nd equation}&\text{y-coefficient, 2nd equation} \end{array}\right]$

which means

$\left[\begin{array}{cc}4&-3\\8&-3\end{array}\right]$

The determinant is computed subtracting diagonals:

$\left | \left[ \begin{array}{cc}a&b\\c&d\end{array}\right]\right | = ad-bc$

So, we have

$\left | \left[\begin{array}{cc}4&-3\\8&-3\end{array}\right] \right | = 4(-3) - 8(-3) = -4(-3) = 12$

8 0

3 years ago

Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu

Alona [7]

Answer:

We have the equation

$c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]$

Then, the augmented matrix of the system is

$\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]$

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

$\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]$

This matrix is in echelon form. Then, now we apply backward substitution:

$8c_4=0\\c_4=0$

$4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0$

$3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0$

$c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0$

Then the system has unique solution that is $(c_1,c_2c_3,c_4)=(0,0,0,0)$ and this imply that the vectors $v_1,v_2,v_3,v_4$ are linear independent.

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3 years ago

Which equation represents the circle described?

bazaltina [42]

We have the equation:

$x^2+y^2-8x-6y+24=0$

By arranging this equation in terms of x and y, we have:

$x^2-8x+y^2-6y=-24 \\ \\$

By using the method of completing the square, we have:

$x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}$