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4 years ago

What’s this problem

Mathematics

1 answer:

Bezzdna [24]4 years ago

3 0

Answer:

use FOIL- Firsts Outsides Insides Lasts

Multiply the -4 and -3 as they are the first numbers of each brackets -4x-3=12
Next the outsides so -4x6i as they are on the outskirts of each brackets -4x6i= -24i
Insides so 3i and -3 as they are in the centre of the two brackets 3ix-3= -9i
Lasts so 3i and 6i as they are at the end of each bracket 3ix6i=18i^2 ( i squared)
add the answers together - 12+-24i+-9i+18i^2= 12-24i-9i+18i^2
Simplify by adding like terms which is -24i and -9i.

<h3>12+-33i+18i^2 is the answer </h3>

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Exercise 3.3.2: Verify that the system x® 0 1 3 3 1 x® has the two solutions 1 1 e 4t and 1 −1 e −2t a) . b) Write down the gene

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The general solution of the system is

$X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\$

Step-by-step explanation:

To verify that the system $X^{'} =\left[\begin{array}{cc}1&3\\3&1\end{array}\right] X$ has the solutions given,

First, we determine the Eigen Values β of the matrix of the form $X^{'} =A X$ Using |A-βI|=0, where I is the Identity Matrix, we have:

$|A-\beta I|=0$

$|\left(\begin{array}{cc}1&3\\3&1\end{array}\right )-\left(\begin{array}{cc}\beta &0\\0&\beta \end{array}\right )|=0$

Subtracting matrices

$\left|\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right |=0$

Taking the determinant

$(1-\beta)(1-\beta)-(3X3)=0\\1-\beta-\beta+\beta^{2}-9=0\\\beta^{2}-2\beta-8=0\\\beta^{2}-4\beta+2\beta-8=0\\\beta(\beta-4)+2(\beta-4)=0\\(\beta-4)((\beta+2)=0\\\beta=4 or -2$

We determine the eigen vector using $\left(\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$

When $\beta$ = -2,

$\left(\begin{array}{cc}3 &3\\3&3 \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$ which implies that $3c_{11}+3c_{12}=0$ and thus

$If c_{11}=1, c_{12}=-1$

When $\beta$ = 4,

$\left(\begin{array}{cc}-3 &3\\3&-3 \end{array}\right)\left(\begin{array}{c}c_{21} &c_{22} \end{array}\right)=0$ which implies that $3c_{21}-3c_{22}=0$ and thus