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3 years ago

How do you simplify -24e5/6e

Mathematics

1 answer:

kifflom [539]3 years ago

4 0

-24e^5/6e = -4e^4
hope it helps

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In the accompanying diagram of ABC ca is extended to D, m∠ABC = 70 and m∠BCA =50 Find m∠DAB

zvonat [6]

Answer:

120 degrees

Step-by-step explanation:

Find the diagram attached.

From the diagram;

Interior angles are m∠BCA and m∠ABC

Exterior angle is m∠DAB

The sum of interior angle of the triangle is equal to exterior

m∠BCA +m∠ABC =m∠DAB

Given

m∠ABC = 70

m∠BCA =50

m∠DAB = 70 + 50

m∠DAB = 120 degrees

Hence the measure of m∠DAB is 120 degrees

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3 years ago

Esta Bond has a big jar of change in his room he has 600 coins total and 240 of them are pennies what percentage of the coins or

KonstantinChe [14]

240/600=0.4
0.4x100= 40
40% of his coins are pennies

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3 years ago

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Triangle RST has vertices located at R (2, 3), S (4, 4), and T (5, 0).

hichkok12 [17]

Answer:im not sure

Step-by-step explanation:

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1 year ago

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mario62 [17]

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3 years ago

Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8

Kitty [74]

Substitute $z=\ln x$ , so that

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)$

Then the ODE becomes

$x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0$
$\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0$

which has the characteristic equation $r^2+1=0$ with roots at $r=\pm i$ . This means the characteristic solution for $y(z)$ is

$y_C(z)=C_1\cos z+C_2\sin z$

and in terms of $y(x)$ , this is

$y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)$

From the given initial conditions, we find

$y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1$
$y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8$

so the particular solution to the IVP is

$y(x)=\cos(\ln x)+8\sin(\ln x)$

4 0

3 years ago