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iragen [17]
3 years ago
12

6x^2=32x+24 Need help learning factoring please put work so I can understand

Mathematics
2 answers:
Stels [109]3 years ago
8 0

Answer:

I do not sorry

Travka [436]3 years ago
7 0

Answer:

2 • (x - 6) • (3x + 2)

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2".

STEP 1 : Equation at the end of step 1

((2•3x2) - 32x) - 24

STEP 2 : Pulling out like terms

Pull out like factors :

6x2 - 32x - 24 = 2 • (3x2 - 16x - 12)

Trying to factor by splitting the middle term

Factoring 3x2 - 16x - 12

The first term is, 3x2 its coefficient is 3 .

The middle term is, -16x its coefficient is -16 .

The last term, "the constant", is -12

Step-1 : Multiply the coefficient of the first term by the constant 3 • -12 = -36

Step-2 : Find two factors of -36 whose sum equals the coefficient of the middle term, which is -16 .

-36 + 1 = -35

-18 + 2 = -16 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -18 and 2

3x2 - 18x + 2x - 12

Step-4 : Add up the first 2 terms, pulling out like factors :

3x • (x-6)

Add up the last 2 terms, pulling out common factors :

2 • (x-6)

Step-5 : Add up the four terms of step 4 :

(3x+2) • (x-6)

Which is the desired factorization

Final result : 2 • (x - 6) • (3x + 2)

Hope this helps~

Smoll mochi

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PLEASE HELP QWQ AsAp with these 4 questions
den301095 [7]

Answer:

Step-by-step explanation:

I can't believe I'm doing this for 5 points, but ok!

For the first 3, we are going to multiply to find the value of that 3 x 3 matrix by picking up the first 2 columns and plopping them down at the end and then multiplying through using the rules for multiplying matrices:

\left[\begin{array}{ccccc}7&4&6&7&4\\-4&8&9&-4&8\\1&8&7&1&8\end{array}\right]  and from there find the sum of the products of the main axes minus the sum of the products of the minor axes, as follows (I'm not going to state the process in the next 2 problems, so make sure you follow it here. This is called the determinate. The determinate is what you get when you evaluate or find the value of a matrix. Just so you know):

(7*8*7)+(4*9*1)+(6*-4*8)-[(1*8*6)+(8*9*7)+(7*-4*4)] which gives us:

392 + 36 - 192 - [48 + 504 - 112] which simplifies to

236 - 440 which is -204

On to the second one:

\left[\begin{array}{ccccc}-8&-4&-1&-8&-4\\1&7&-3&1&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-8*7*9)+(-4*-3*8)+(-1*1*9)-[(8*7*-1)+(9*-3*-8)+(9*1*-4)] which gives us:

-504 + 96 - 9 - [-56 + 216 - 36] which simplifies to

-417 - 124 which is -541, choice c.

Now for the third one:

\left[\begin{array}{ccccc}-2&-2&-5&-2&-2\\2&7&-3&2&7\\8&9&9&8&9\end{array}\right] and multiplying gives us

(-2*7*9)+(-2*-3*8)+(-5*2*9)-[(8*7*-5)+(9*-3*-2)+(9*2*-2)] which gives us:

-126+48-90-[-280+54-36] which simplifies to

-168 - (-262) which is 94, choice c again.

Now for the last one. I'll show you the set up for the matrix equation; I solved it using the inverse matrix. So I'll also show you the inverse and how I found it.

\left[\begin{array}{cc}-4&-5&\\-6&-8\\\end{array}\right] \left[\begin{array}{c}x\\y\\\end{array}\right] = \left[\begin{array}{c}-5\\-2\\\end{array}\right] and I found the inverse of the 2 x 2 matrix on the left.

Find the inverse by:

* finding the determinate

* putting the determinate under a 1

* multiply that by the "mixed up matrix (you'll see...)

First things first, the determinate:

|A| = (-4*-8) - (-6*-5) which simplifies to

|A| = 32 - 30 so

|A| = 2; now put that under a 1 and multiply it by the mixed up matrix. The mixed up matrix is shown in the next step:

\frac{1}{2}\left[\begin{array}{cc}-8&5\\6&-4\end{array}\right]  (to get the mixed up matrix, swap the positions of the numbers on the main axis and then change the signs of the numbers on the minor axis). Now we multiply in the 1/2 to get the inverse:

\left[\begin{array}{cc}-4&\frac{5}{2}\\3&-2\\\end{array}\right] Multiply that inverse by both sides of the equation. This inverse "undoes" the matrix that's already there (like dividing the matrix that's already there by itself) which leaves us with just the matrix of x and y. Multiply the inverse matrix by the solution matrix:

\left[\begin{array}{c}x&y\end{array}\right] =\left[\begin{array}{cc}-4&\frac{5}{2} \\3&-2\end{array}\right] *\left[\begin{array}{c}-5&-2\\\end{array}\right] and that right side multiplies out to

x = 20 - 5 which is

x = 15 and

y = -15 + 4 which is

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(It works, I checked it)

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3 years ago
Please help!!
Kamila [148]

The relationship between the trigonometric ratios are cos(X) = cos(A), cos(Y) = cos(C) and cos(Z) = cos(B)

<h3>The relationship between the trigonometric ratios</h3>

The given parameters are:

  • Scale factor of dilation, k = 2
  • cos(X) = 2.5/5/59

Since the transformation is a dilation, it means that both triangles are similar.

So, we have:

cos(X) = cos(A), cos(Y) = cos(C) and cos(Z) = cos(B)

<h3>The segments AC and AB. </h3>

In (a), we have:

Both triangles are similar.

This means that the corresponding points are:

X and A

Y and  C

Z and B

So, we have:

AC = 2 * XY

AB  = 2 * XZ

Read more about dilation at:

brainly.com/question/13176891

#SPJ1

4 0
2 years ago
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