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3 years ago

Help pleaseeeeeeeeeee

Mathematics

1 answer:

avanturin [10]3 years ago

5 0

I believe your answer is going to be increasing linear. If it's doubling constantly then it's increasing so that takes out 2 options. And since it's constant, it isn't exponential, therefore you only have increasing linear left.

I hope I was able to help. Best of luck.

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Mr. Simpson deposits his money in a savings account at the Springfield Bank. Would he earn more money with simple interest or wi

Neko [114]

Your Anwser is Compound Interest

4 0

3 years ago

The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Sav [38]

Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then

$\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v$

$\implies\ln|u|=\ln|v|+C$

$\implies\ln(y^2+4)=\ln|\ln x|+C$

$\implies y^2+4=e^{\ln|\ln x|+C}$

$\implies y^2=C|\ln x|-4$

$\implies y=\pm\sqrt{C|\ln x|-4}$

4 0

3 years ago

Solve the system of equations.

Xelga [282]

Answer:

b. x=1, y=2, z=3

Step-by-step explanation:

The system of equations ...

3x +2y +z = 10
9x -6y +z = 0
x -y -3z = -10

has solution (x, y, z) = (1, 2, 3) . . . . matches choice B.

_____

While it is convenient to solve this using a graphing calculator or web site, one can easily solve the system by hand.

Subtract the second equation from 3 times the first:

3(3x +2y +z) -(9x -6y +z) = 3(10) -(0)

12y + 2z = 30 . . . . simplify

Dividing this result by 2 gives ...

6y +z = 15 . . . . . . [eq4]

Subtract 3 times the third equation from the first:

(3x +2y +z) -3(x -y -3z) = (10) -3(-10)

5y +10z = 40 . . . . simplify

y + 2z = 8 . . . . . . . divide by 5 . . . . . [eq5]

The two equations [eq4] and [eq5] can be solved any of the ways you usually solve two equations in two variables. Here, we'll use the first equation to write an expression for z that we can substitute into the second equation.

z = 15 -6y . . . . . subtract 6y from [eq4]

y + 2(15 -6y) = 8 . . . . . substitute for z in [eq5]

-11y +30 = 8 . . . . . simplify

-11y = -22 . . . . . . . subtract 30

y = 2 . . . . . . . . . . . divide by the coefficient of y

z = 15 -6(2) = 3 . . . . substitute for y in our equation for z

Substituting these values for y and z into the third original equation gives ...

x - 2 -3(3) = -10

x -11 = -10 . . . . . . . . simplify

x = 1 . . . . . . . . . . . . add 11

The solution to the above system of equations is (x, y, z) = (1, 2, 3).

_____

<em>Comment on the problem statement</em>

Math is generally unforgiving of imprecision. The given system of equations has no variable "z", and some other typos are apparently involved. That is why we rewrote the system to the equations shown above.

It is very easy to mistake z for 2, or g for 9, or o for 0, or 1 for 7. There are other confusions that are possible, as well. Letters I (eye) and l (ell) are easily confused, and may be confused with 1 (one) as well. Sometimes y and 4, or 4 and 9, can also be written so as to be difficult to tell apart. Great care must be taken when handwriting these symbols.

7 0

4 years ago

Someone give me the answers to this

strojnjashka [21]

Range: the set of all y-values of a graph or equation.

Y-intercept: the y-value where the graph intersects the y-axis.

Domain: the set of all x-values of a graph or equation.

X-intercept: the x-values where the graph intersects the x-axis.

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

Using only the digits 0 -9 enter a postive vaklue for x

Ghella [55]

Answer:

math w a y .com

Step-by-step explanation:

3 0

3 years ago