Why isn't anyone answering this?

Answer asap please

sergeinik [125]

Answer:

t = 0

If the loan is to be paid in 0 months, the amount is very large

m = 50

Of the loan is to be paid in an indefinite amount of time, monthly payment would be 50

Step-by-step explanation:

(12000 + 600t)/12t

1000/t + 50

Asymptotes:

t = 0

If the loan is to be paid in 0 months, the amount is very large

m = 50

Of the loan is to be paid in an indefinite amount of time, monthly payment would be 50

3 0

3 years ago

Which equation could represent the graph below?

coldgirl [10]

Equation 2. If you look at the graph, you see that the slope is negative 3/4 because the rise over run is -3/4. The only equation with the slope -3/4 is 2, so the answer is 2.

8 0

3 years ago

How do I solve this? help please! xx

Anna35 [415]

This graph is composed of four straight line segments. You'll need to determine the slope, y-intercept and domain for each of them. Look at the first segment, the one on the extreme left. Verify yourself that the slope of this line segment is 1 and that the y-intercept would be 0 if you were to extend this segment all the way to the y-axis. Thus, the rule (formula, equation) for this line segment would be f(x)=1x+0, or just f(x)=x, for (-3,-1). Use a similar approach to write rules for the remaining three line segments.

Present your answer like this:
x, (-3,-1)
f(x) = -1, (-1,0)
one more here
one more here

5 0

3 years ago

can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

6 0

3 years ago

Solve for m the problem is y=mx+b

lana66690 [7]

It would be -b/x+y/x

4 0

3 years ago

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