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2 years ago

M*(x+y)+4y/(x^2-xy-y^2)=(x+2y)/(x^2-xy-Y^2),find the value of M

Mathematics

1 answer:

dybincka [34]2 years ago

3 0

Answer: M = $\frac{x-2y}{x+y}$

Step-by-step explanation:

$\frac{M(x+y)+4y}{x^{2}-xy-y^{2} } =\frac{(x+2y)}{x^{2} -xy-y^{2} }$

M (x + y) + 4y = x + 2y (Multiply $x^{2} - xy-y^{2}$ in both sides)

M (x + y) = x + 2y - 4y

M (x + y) = x - 2y (Divide (x+y) from both sides)

M = $\frac{x-2y}{x+y}$

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3 years ago

The sum of three consecutive integers is 90. Find the three numbers by letting x

Law Incorporation [45]

Answer:

x = 28

Step-by-step explanation:

let us take three consecutive numbers be (x + 1), (x + 2) and (x + 3)

Now,

The sum of three consecutive integers is 90

i.e.,

(x + 1) + (x + 2) + (x + 3) = 90

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x = 28Thus, The value of x is 28

<h2><em><u>VERIFICATION</u></em><em><u>:</u></em></h2>

(x + 1) + (x + 2) + (x + 3) = 90

(28 + 1) + (28 + 2) + (28 + 3) = 90

28 + 1 + 28 + 2 + 28 + 3 = 90

90 = 90

8 0

3 years ago

Read 2 more answers

Check answer please

Cerrena [4.2K]

The fourth or the D) Option is correct.

To find the new induced matrix via a scalar quantified multiplication we have to multiply the scalar quantity with each element surrounded and provided in a composed (In this case) 3×3 or three times three matrix comprising 3 columns and 3 rows for each element which is having a valued numerical in each and every position.

Multiply the scalar quantity with each element with respect to its row and column positioning that is,

Row × Column. So;

(1 × 1) × 7, (2 × 1) × 7, (3 × 1) × 7, (1 × 2) × 7, (2 × 2) × 7, (3 × 2) × 7, (1 × 3) × 7, (2 × 3) × 7 and (3 × 3) × 7. This will provide the final answer, that is, the D) Option.

To interpret and make it more interesting in LaTeX form. Here is the solution with LaTeX induced matrix.

$\mathcal{A = \begin{bmatrix}1 & 0 & 3 \\ 2 & -1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}$

$\mathbf{\therefore \quad 7A = 7 \times \begin{bmatrix}1 & 0 & 3 \\ 2 & - 1 & 2 \\ 0 & 2 & 1 \\ \end{bmatrix}}$

$\mathbf{\therefore \quad \begin{bmatrix}7 \times 1 & 7 \times 0 & 7 \times 3 \\ 7 \times 2 & 7 \times -1 & 7 \times 2 \\ 7 \times 0 & 7 \times 2 & 7 \times 1 \\ \end{bmatrix}}$

$\therefore \quad \begin{\bmatrix}7 & 14 & 0 \\ 0 & -7 & 14 \\ 21 & 14 & 7 \end{bmatrix}$

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