PLEASE HELP EASY MATH PLEASE ASAP

Mathematics

1 answer:

Sedbober [7]2 years ago

4 0

Answer:

Step-by-step explanation:

I don't really know what your question is asking for but I can tell you there's a quadratic formula that solves for x-ints. There's a quadratic equation which is Ax + bx + c = 0.

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Expand. Your answer should be a polynomial in standard form. (3c + 2)(c^2-6c-4)

evablogger [386]

Answer:

After expanding the polynomial $(3c + 2)(c^2-6c-4)$ we get $3c^3-16c^2-24c-8$

Step-by-step explanation:

We need to expand the polynomial $(3c + 2)(c^2-6c-4)$

Multiply the terms:

$(3c + 2)(c^2-6c-4)\\=3c(c^2-6c-4)+2(c^2-6c-4)\\=3c^3-18c^2-12c+2c^2-12c-8\\=3c^3-18c^2+2c^2-12c-12c-8\\=3c^3-16c^2-24c-8$

So, after expanding the polynomial $(3c + 2)(c^2-6c-4)$ we get $3c^3-16c^2-24c-8$

3 0

2 years ago

Find the? inverse, if it? exists, for the given matrix. [4 3] [3 6]

True [87]

Answer:

Therefore, the inverse of given matrix is

$=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}$

Step-by-step explanation:

The inverse of a square matrix $A$ is $A^{-1}$ such that

$A A^{-1}=I$ where I is the identity matrix.

Consider, $A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]$

$\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}$

$\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0$

$\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}$