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3 years ago

Which table represents a linear function?

Mathematics

1 answer:

Luda [366]3 years ago

3 0

A because is A jajajjajajjajajajajjajjaj

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Write 2^60 as an exponent with a base of 8

JulijaS [17]

Answer: 2^60 = 8^20

Step-by-step explanation: 8=2^3

(2^3)^20 Multiply the exponents

3 0

3 years ago

Read 2 more answers

Solve the equation x^3 y^'" + 5x^2 y" + 7xy' + 8y = 0

seraphim [82]

Make the substitution $t=\ln x$ , then compute the derivatives of $y$ with respect to $t$ via the chain rule.

First derivative

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}$

Second derivative

Let $f(t)=\frac{\mathrm dy}{\mathrm dt}$ .

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac fx\right]=\dfrac{x\frac{\mathrm df}{\mathrm dx}-f}{x^2}$

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}$

$\implies\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}\right)$

Third derivative

Let $g(t)=\frac{\mathrm df}{\mathrm dt}=\frac{\mathrm d^2y}{\mathrm dt^2}$ .

$\dfrac{\mathrm d^3y}{\mathrm dx^3}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{g-f}{x^2}\right]=\dfrac{x^2\left(\frac{\mathrm dg}{\mathrm dx}-\frac{\mathrm df}{\mathrm dx}\right)-2x(g-f)}{x^4}$

$\dfrac{\mathrm dg}{\mathrm dx}=\dfrac{\mathrm dg}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^3y}{\mathrm dt^3}$

$\implies\dfrac{\mathrm d^3y}{\mathrm dx^3}=\dfrac{x^2\left(\frac1x\frac{\mathrm dg}{\mathrm dt}-\frac1x\frac{\mathrm df}{\mathrm dt}\right)-2x(g-f)}{x^4}=\dfrac1{x^3}\left(\dfrac{\mathrm d^3y}{\mathrm dt^3}-3\dfrac{\mathrm d^2y}{\mathrm dt^2}+2\dfrac{\mathrm dy}{\mathrm dt}\right)$

Substituting $y(t)$ and its derivatives into the ODE gives a new one that is linear in $t$ :

$\left(\dfrac{\mathrm d^3y}{\mathrm dt^3}-3\dfrac{\mathrm d^2y}{\mathrm dt^2}+2\dfrac{\mathrm dy}{\mathrm dt}\right)+5\left(\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}\right)+7\dfrac{\mathrm dy}{\mathrm dt}+8y=0$

$\dfrac{\mathrm d^3y}{\mathrm dt^3}+2\dfrac{\mathrm d^2y}{\mathmr dt^2}+4\dfrac{\mathrm dy}{\mathrm dt}+8y=0$

$y'''+2y''+4y'+8y=0$

which has characteristic equation

$r^3+2r^2+4r+8=(r+2)(r^2+4)=0$

with roots $r=-2$ and $r=\pm2i$ , so that the characteristic solution is

$y_c(t)=C_1e^{-2t}+C_2\cos2t+C_3\sin2t$

Replace $t=\ln x$ to solve for $y(x)$ :

$y_c(x)=C_1e^{-2\ln x}+C_2\cos(2\ln x)+C_3\sin(2\ln x)$

$\boxed{y(x)=\dfrac{C_1}{x^2}+C_2\cos(2\ln x)+C_3\sin(2\ln x)}$

3 0

2 years ago

The shape is a rhombus if and only if the diagonals are perpendicular and the sides are congruent

harina [27]

Answer:

The statement is True

Step-by-step explanation:

Rhombus is a quadrilateral with the following characteristics;

All sides are congruent by definition.
The diagonals bisect the angles.
The diagonals are perpendicular bisectors of each other.
Adjacent angles are supplementary.
All the four sides are equal.

5 0

3 years ago

And, what is the least common multiple of 6,15, and 5

iren2701 [21]

Answer:

there is no multiplyer becuase 1 dosnt work no 2 nor3 nor4 nor5 nor6 and then you will be greater than 6 so it dosnt fit into 5 so

????/

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

What is the product of −2 1/4 and −4 1/2 ?

frozen [14]

To multiply fracti0ons, we need to convert them to improper fractions:

-2 1/4 = -9/4
-4 1/2 = -9/2

Now, lets multiply:

-9/4 * -9/2 = 81/8

Hope this helps!

Take note that the answer is positive because when multiplying, two negatives make a positive.

4 0

3 years ago