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2 years ago

For what values of x and y is the equation (x2 + y2)2 = (x2 – y2)2 + (2xy)2 true?

Mathematics

1 answer:

denis-greek [22]2 years ago

6 0

Answer:

A) all values of x, and all values of y

B) (3a + 4b) (9a² − 12ab + 16b²)

Step-by-step explanation:

(x² + y²)² = (x² − y²)² + (2xy)²

x⁴ + 2x²y² + y⁴ = x⁴ − 2x²y² + y⁴ + 4x²y²

x⁴ + 2x²y² + y⁴ = x⁴ + 2x²y² + y⁴

27a³ + 64b³

(3a)³ + (4b)³

(3a + 4b) (9a² − 12ab + 16b²)

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If f(x) = 2(x)^2+5 sqrt (x-2). complete the following statement. <br><br> f(2)=_____

dangina [55]

Answer:

The value of function at x = 2 is 8.

Step-by-step explanation:

Given : Function $f(x)=2x^2+5\sqrt{x-2}$

We have to find the value of given function at x = 2 , that is f(2)

Consider the given function $f(x)=2x^2+5\sqrt{x-2}$

To find the value f(2) put x = 2 in given function ,

We have,

$f(2)=2(2)^2+5\sqrt{2-2}$

Simplify, we have,

$f(2)=8+5\sqrt{0}$

Simplify further, we have,

$f(2)=8$

Thus, the value of function at x = 2 is 8.

4 0

3 years ago

Read 2 more answers

(1 point) (a) Find the point Q that is a distance 0.1 from the point P=(6,6) in the direction of v=⟨−1,1⟩. Give five decimal pla

natima [27]

Answer:

following are the solution to the given points:

Step-by-step explanation:

In point a:

$\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}$

$=\sqrt{1+1}\\\\=\sqrt{2}$

calculating unit vector:

$\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}$

the point Q is at a distance h from P(6,6) Here, h=0.1

$a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071$

the value of Q= (5.92928 ,6.07071 )

In point b:

Calculating the directional derivative of $f (x, y) = \sqrt{x+3y}$ at P in the direction of $\vec{v}$

$f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\$

$=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557$

$\vec{v} = 1.97651557$

In point C:

Computing the directional derivative using the partial derivatives of f.

$f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}$