The difference between two numbers is 14 and the difference between their squares is 448, then find the numbers.

Mathematics

2 answers:

klio [65]2 years ago

8 0

Answer:

23, 9

Step-by-step explanation:

Let a, b - be the numbers.

$a-b = 14\\a^2-b^2=448 = (a-b)(a+b) = 448\\(14)(a+b)=448\\a+ b = 448/14 = 32$

$a - b = 14\\a + b = 32\\2a = 46\\a = 23 => b = 9$

AlladinOne [14]2 years ago

4 0

Answer with Step-by-step explanation:

Let a, b be the two numbers.

The difference between two numbers is 14

i.e. a-b=14 -------(1)

The difference between their squares is 448

i.e.

$a^2-b^2=448$

i.e. (a-b)(a+b) = 448

(14)(a+b)=448

a+ b = 448/14

a+b = 32 --------(2)

Adding equation (1) and (2)

a-b+a+b=14+32

2a = 46

a= 23

Putting a=23 in a+b=32, we get

23+b=32

b=32-23

b= 9

Hence, two numbers are:

23 and 9

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A= [1, 3; 2, 1], B=[3, 6; -1, 1]. Find AB & BA if possible

steposvetlana [31]

Answer:

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Step-by-step explanation:

For two matrix P and Q, the product, say PQ is defined when:

The number of columns of P = The number of rows of Q

Since A is a 2×2 matrix and B is also a 2×2 matrix

Thus both AB and BA are possible.

So AB is:

$AB\Rightarrow\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\\AB\Rightarrow\quad \begin{bmatrix}3\times 1+3\times (-1) & 6\times 1+3\times 1\\3\times 2+1\times (-1) & 6\times 2+1\times 1\end{bmatrix}\\AB\Rightarrow \quad \begin{bmatrix}0 & 9\\ 5 & 13\end{bmatrix}$

BA is:

$BA\Rightarrow\begin{bmatrix}3 & 6\\ -1 & 1\end{bmatrix}\begin{bmatrix}1 & 3\\ 2 & 1\end{bmatrix}\\BA\Rightarrow\quad \begin{bmatrix}3\times 1+6\times 2 & 3\times 3+6\times 1\\(-1)\times 1+1\times 2 & (-1)\times 3+1\times 1\end{bmatrix}\\BA\Rightarrow \quad \begin{bmatrix}15 & 15\\ 1 & -2\end{bmatrix}$