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3 years ago

Graphs of the following equations are straight lines expect?

Mathematics

1 answer:

natka813 [3]3 years ago

7 0

I think you would do that

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To go to the state park. Mary's family had to pay $3 for the car. plus $2 for each person in the car. Write an expression that c

Helga [31]

Answer:

3+2x or 2x+3

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Does the equation specify a function with independent variable x? If so, find the domain of the function. If not, find a valu

drek231 [11]

Yes, the equation specifies a function with independent variable x.

Equation: 6x + 19y = 7

This can be put in y = mx + b form,

6x + 19y = 7

19y = -6x + 7

y = - $\frac{6}{19}$ x + $\frac{7}{19}$

Since the equation can be put in the form of a linear function, the domain is all read numbers (-∞,∞) or -∞ < x < ∞ .

7 0

3 years ago

Consider xưy" – 14xy' + 56y = 0. Find all values of r such that y=x" satisfies the differential equation for x > 0. Enter as

beks73 [17]

I bet the ODE is supposed to read

$x^2y''-14xy'+56y=0$

Then if $y=x^r$ , we have $y'=rx^{r-1}$ and $y''=r(r-1)x^{r-2}$ , and substituting these into the ODE gives

$r(r-1)x^r-14rx^r+56x^r=0\implies r(r-1)-14r+56=r^2-15r+56=0$

Solving for r, we find

$r^2-15r+56=(r-8)(r-7)=0\implies \boxed{r=8\text{ or }r=7}$

so that $y_1=x^8$ and $y_2=x^7$ are two fundamental solutions to the ODE. Thus the general solution is

$\boxed{y(x)=C_1x^8+C_2x^7}$

Given that $y(1)=4$ and $y'(1)=3$ , we get

$\begin{cases}4=C_1+C_2\\3=8C_1+7C_2\end{cases}\implies C_1=-25\text{ and }C_2=29$

So the particular solution is

$\boxed{y(x)=29x^7-25x^8}$

4 0

4 years ago

What is the solution to the system of equations? x+y+3z=-4 2x-z=-3 -x-y-2z=5

goblinko [34]

x=-4-y-3z

x=-3/2+1/2z

x=-5-y-2z

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3 years ago

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