All categories

3 years ago

Solving special systems.

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

8 0

Answer:

Step-by-step explanation:

it has no solutions b/c the lines are parallel...but at a different offset.. so they never cross

You might be interested in

The parabola y=x^2 is shifted up by 8 units. What is the new equation of the parabola?

elixir [45]

Answer:

y=x^2+8

Step-by-step explanation:

5 0

2 years ago

Peggy was taking a timed 90-question test. She was able to correctly answer 80% of the questions but had to skip the rest. How m

professor190 [17]

Multiply your total by your percentage

90 * .80 = 72

72 is the number of questions she answered correctly

7 0

3 years ago

Which expressions are equivalent to 3x

musickatia [10]

x is powering both numbers so it can be outside the parenthesis.

We have given that 3^x.

<h3>What is the expression?</h3>

An expression or mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context.

The first one isn't an answer because 3^x is exponential while x^3 is a cubic function.

If you draw them you will see that they are very different.

B is correct because we can divide both numerator and denominator by 6 and we get 3^x.

C is not correct because x is not powering 3 so we cannot divide both by 6D is correct because 3^(x-1) is the same as

and when multiplied by 3 we get 3^x

3^x*3^(-1) = 3^x/3

E is not correct.

will understand after the explanation in DF is correct.

x is powering both numbers so it can be outside the parenthesis.

The question is incomplete the complete question is,

Which expressions are equivalent to the one below? Check all that apply.

3^x

A. x^3

B.(18/6)^x

C.18^x/3

D.3(3^(x-1))

E.3(3^(x+1))

F.18^x/6^x

To learn more about the expression visit:

brainly.com/question/723406

#SPJ1

7 0

1 year ago

Write the equation for the following relation. Include all of your work in your final answer. Submit your solution.

ozzi

Answer:

f(x) = x³ x>1

Step-by-step explanation:

2 8=2³

3 27=3³

4 64=4³

5 125=5³

f(x) = x³ x>1

4 0

2 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