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Anika [276]
2 years ago
9

What is the domain of f(x)=5x-7

Mathematics
2 answers:
Kobotan [32]2 years ago
8 0

Answer:

(-infinity, infinity)

Step-by-step explanation:

You’re welcome, sorry I couldn't out the symbols. I don't have them on my keyboard :).

KonstantinChe [14]2 years ago
4 0

Answer:

R (all real numbers)

R (all real numbers)

Step-by-step explanation:

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Necesito esto para hoy y no entiendo
BaLLatris [955]

Operación: (+6)-(3). Piso Final: +3

Inicial: +8 Operación: (+8)+(+1) Piso final: +9

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3 0
2 years ago
2<br> 2. Evaluate (a + y)+ 2y if a = 5 and y = -3.<br> F 58<br> G -2<br> H 70<br> J 10
NARA [144]
<h3>Answer:  G) -2</h3>

=======================================================

Explanation:

I'm assuming you meant to say (a+y)^2 + 2y

Replace each copy of 'a' with 5. Replace each copy of 'y' with -3. Use PEMDAS to simplify.

(a+y)^2 + 2y

(5 + (-3))^2 + 2(-3)

(5-3)^2 + 2(-3)

(2)^2 + 2(-3)

4 + 2(-3)

4 - 6

-2

So (a+y)^2 + 2y = -2 when a = 5 and y = -3.

3 0
2 years ago
A rectangular solid and a cube are of equal volume. If the rectangular solid is 20 centimeters long, 5 centimeters wide, and 10
abruzzese [7]

Answer:

10 centimeters.

Step-by-step explanation:

First, we need to remember what's the formula to get the volume of a rectangular solid and a cube.

The volume of the first equals:

Volume = Length x Width x Height

While the volume of the cube is:

Volume = a^{3} where a is the edge.

We are given the measures of the rectangular solid so we can calculate its volume:

Volume= (20)(5)(10)=1000 cubic cms.

Now, we know that  both the volume of the rectangular solid and the cube are the same so we will use this information to calculate the edge of the cube.

1000=a^3 \\\sqrt[3]{1000} =\sqrt[3]{a^3} \\10=a

Thus the length of an edge of the cube is 10 centimeters

5 0
2 years ago
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
Please help !!!!!!!!!!!!!!!!!!!!!!!!!!!You survey 127 employees, and learn that the number one thing on their wish list is free
valina [46]

Answer:

c

Step-by-step explanation:

3 0
3 years ago
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