Simplify in standard form <br> 8+5i/5+2i

Please help, i’m so behind on my work :(( image attached

gregori [183]

Answer:

5/8

Step-by-step explanation:

28x + 4 can be factorised because 28 is a multiple of 4

so it can become

4(7x+1)

35x + 5 can be also factorised because 35 is a multiple of 5

5(7x+1)

we can now rewrite the expression as

6/4(7x+1) * 5(7x+1)/12

we can “eliminate” the two 7x+1 and simplify 6 with 12

so it remains

1/4 * 5/2 = 5/8

8 0

3 years ago

What are the coordinates of the point (-1, 6) after a reflection across the x-axis?

maks197457 [2]

Answer:

B(-1,-6)

Step-by-step explanation:

reflecting reflecting over the x axis changes the y outputs.

4 0

3 years ago

Read 2 more answers

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

$(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0$

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

2 years ago

What is the factor of 12 but not the multiple of 3

algol13

The factor of 12 is 4

7 0

3 years ago

F(x)=x+7<br> f(-1)<br> help!

DIA [1.3K]

Answer:

f(-1) = 6

Step-by-step explanation:

f(x) = x + 7 when x = -1

~Substitute

= (-1) + 7

~Simplify

= 6

Best of Luck!

7 0

2 years ago

Simplify in standard form 8+5i/5+2i