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2 years ago

The diagram shows a cuboid 4cm, 5cm and 9cm what is the surface area of the cuboid?

Mathematics

2 answers:

larisa86 [58]2 years ago

5 0

280cm because if you multiply all of them up

DIA [1.3K]2 years ago

4 0

<h3>given:</h3>

length= 9 cm

width= 5 cm

height= 4 cm

the surface area of the given cuboid.

<h3>solution:</h3>

$s.a = 2(lb + bh + lh)$

$s.a = 2((9 \times 5) + (5 \times 4) + (9 \times 4))$

$s.a = 2(45 + 20 + 36)$

$s.a = 2 \times 101$

$s.a = 202 \: {cm}^{2}$

therefore, the surface area of the given cuboid is 202 square centimeters.

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How do you do (a) and (b)?

bulgar [2K]

Answer:

See solution below

Step-by-step explanation:

(a) If n=0 or 1, the equation

(1) $y' = a(t)y + f(t)y^n$

would be a simple linear differential equation. So, we can assume that n is different to 0 or 1.

Let's use the following substitution:

(2) $z=y^{n-1}$

Taking the derivative implicitly and using the chain rule:

(3) $z'=(1-n)y^{-n}y'$

Multiplying equation (1) on both sides by

$(1-n)y^{-n}$

we obtain the equation

$(1-n)y^{-n}y' = (1-n)y^{-n}a(t)y+(1-n)y^{-n}f(t)y^n$

reordering:

$(1-n)y^{-n}y' = (1-n)y^{-n}ya(t)+(1-n)y^{-n}y^nf(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)y^{0}f(t)$

$(1-n)y^{-n}y' = (1-n)y^{1-n}a(t)+(1-n)f(t)$

Now, using (2) and (3) we get:

$z'= (1-n)za(t)+(1-n)f(t)$

which is an ordinary linear differential equation with unknown function z(t).

(b)

The equation we want to solve is

(4) $xy'+ y = x^4 y^3$

Here, our independent variable is x (instead of t)

Assuming x different to 0, we divide both sides by x to obtain:

$y'+\frac{1}{x}y = x^3 y^3$

$y' = -\frac{1}{x}y+x^3 y^3$

Which is an equation of the form (1) with

$a(x)=-\frac{1}{x}$

$f(x)=x^3$

$n=3$

So, if we substitute

$z=y^{-2}$

we transform equation (4) in the lineal equation

(5) $z'=\frac{2}{x}z-2x^3$

and this is an ordinary lineal differential equation of first order whose

integrating factor is

$e^{\int (-\frac{2}{x})dx}$

but

$e^{\int (-\frac{2}{x})dx}=e^{-2\int \frac{dx}{x}}=e^{-2ln(x)}=e^{ln(x^{-2})}=x^{-2}=\frac{1}{x^2}$

Similarly,

$e^{\int (\frac{2}{x})dx}=x^2$

and the general solution of (5) is then

$z(x)=x^2\int (\frac{-2x^3}{x^2})dx+Cx^2=-2x^2\int xdx+Cx^2=\\\ -2x^2\frac{x^2}{2}+Cx^2=-x^4+Cx^2$

where C is any real constant

Reversing the substitution

$z=y^{-2}$

we obtain the general solution of (4)

$y=\sqrt{\frac{1}{z}}=\sqrt{\frac{1}{-x^4+Cx^2}}$

Attached there is a sketch of several particular solutions corresponding to C=1,4,6

It is worth noticing that the solutions are not defined on x=0 and for C<0

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