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patriot [66]
3 years ago
10

(2x-3)(3x+4) expand and simplify

Mathematics
1 answer:
elixir [45]3 years ago
5 0

Answer:

6x² - x - 12

Step-by-step explanation:

Each term in the second factor is multiplied by each term in the first factor.

(2x - 3)(3x + 4)

= 2x(3x + 4) - 3 (3x + 4) ← distribute both parenthesis

= 6x² + 8x - 9x - 12 ← collect like terms

= 6x² - x - 12 ← in expanded form


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What is 78.74 rounded to the nearest tenth?
sveticcg [70]

Answer:

The answer is <u>80</u>

Step-by-step explanation:

78.74 is closer to 80 than it is to 70.


4 0
3 years ago
Read 2 more answers
Orthogonally diagonalize the​ matrix, giving an orthogonal matrix P and a diagonal matrix D. To save​ time, the eigenvalues are
alexgriva [62]

Answer:

P=\left(\begin{array}{ccc}-\frac{2}{3}&-\frac{2}{3}&\frac{1}{3}\\\frac{1}{\sqrt{5}}&0&\frac{2}{\sqrt{5}}\\-\frac{4}{3\sqrt{5}}&\frac{\sqrt{5}}{3}&\frac{2}{3\sqrt{5}}\end{array}\right)

Step-by-step explanation:

It is a result that a matrix A is orthogonally diagonalizable if and only if A is a symmetric matrix.  According with the data you provided the matrix should be

A=\left(\begin{array}{ccc}-9&-4&2\\ -4&-9&2\\2&2&-6\\\end{array}\right)

We know that its eigenvalues are \lambda_{1}=-14, \lambda_{2}=-5, where \lambda_{2}=-5 has multiplicity two.

So if we calculate the corresponding eigenspaces for each eigenvalue we have

E_{\lambda_{1}=-14}=\langle(-2,-2,1)\rangle,E_{\lambda_{2}=-5}=\langle(1,0,2),(-1,1,0)\rangle..

With this in mind we can form the matrices P, D that diagonalizes the matrix A so.

P=\left(\begin{array}{ccc}-2&-2&1\\1&0&2\\-1&1&0\\\end{array}\right)

and

D=\left(\begin{array}{ccc}-14&0&0\\0&-5&0\\0&0&-5\\\end{array}\right)

Observe that the rows of P are the eigenvectors corresponding to the eigen values.

Now you only need to normalize each row of P dividing by its norm, as a row vector.

The matrix you have to obtain is the matrix shown below

3 0
3 years ago
2. Which of the following is the equation of a circle with the radius of 1.5 and its center at (-3,2)?
Iteru [2.4K]
(x - xo)^2 + (y - yo)^2 = r^2

[x - (-3)]^2 + [y - (2)]^2 = (1.5)^2

(x+3)^2 + (y-2)^2 = 2.25

Answer: option C

8 0
3 years ago
How do you integrate arctan(x dx? i think that if you simplify the integral you get:?
Temka [501]
Integration by parts will help here. Letting u=\arctan x and \mathrm dv=\mathrm dx, you end up with \mathrm du=\dfrac{\mathrm dx}{1+x^2} and v=x. Now

\displaystyle\int\arctan x\,\mathrm dx=uv-\int v\,\mathrm du
\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\int\frac x{1+x^2}\,\mathrm dx

For the remaining integral, setting y=1+x^2 gives \dfrac{\mathrm dy}2=x\,\mathrm dx, so

\displaystyle\int\frac x{1+x^2}\,\mathrm dx=\frac12\int\frac{\mathrm dy}y=\frac12\ln|y|+C=\frac12\ln(1+x^2)+C

Putting everything together, you end up with

\displaystyle\int\arctan x\,\mathrm dx=x\arctan x-\frac12\ln(1+x^2)+C
6 0
3 years ago
Please answer ASAP! Thank you in advance, and remember to wash your hands consistently for 1 minute.
Genrish500 [490]
W=144 I think :). Hope I helped!
3 0
3 years ago
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