All categories

3 years ago

What is 3x + 5y ????

1 answer:

6 0

3x + 5y
Equal it out to zero.
3x + 5y = 0
Lets solve for x first.
3x = -5y
Divide by 3 to get x by itself.
x = -5y/3
Now solve for y.
3x + 5y = 0
Add -3 to both sides.
5y = -3x
Divide both sides by 5.
y = -3/5x

I Sure Hope This Helps!

You might be interested in

Which of the following is a right triangle?

natali 33 [55]

Answer:

Step-by-step explanation:

the right choice is C

you use c^2=a^2+b^2

where the largest side is the hypotenuse

8 0

2 years ago

Read 2 more answers

Order these numbers from least to greatest.<br> 3.195, 67/20, 3.19, 3 1/3

galina1969 [7]

The numbers from least to greatest are 3.19, 3.195, 3 1/3, and 67/20. This is because 67/20 is equivalent to 3.35 in decimal form, which is greater than all of the other numbers, including 3 1/3, which is 3.3333.

4 0

2 years ago

What is the slope of this line of best fit?

Bas_tet [7]

Answer:

The slope of the line of best fit is $\frac{-6}{7}$ ⇒ 2nd option

Step-by-step explanation:

The formula of the slope of a line is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

<em>To find the slope of the best line fit choose two points their positions make the number of the points over the line equal to the number of the points below the line</em>

From the attached graph points (1 , 9) and (8 , 3) are the best choice

∵ The line passes through points (1 , 9) and (8 , 3)

∴ $x_{1}$ = 1 and $x_{2}$ = 8

∴ $y_{1}$ = 9 and $y_{2}$ = 3

- Substitute them in the formula of the slope

∴ $m=\frac{3-9}{8-1}=\frac{-6}{7}$

∴ The slope of the line of best fit is $\frac{-6}{7}$

6 0

2 years ago

Use the information given to enter an equation in standard form.

nexus9112 [7]

Answer:

y - 1.4x = - 6.8

Step-by-step explanation:

To find the equation: find the slope, use point-slope form to calculate the equation, then convert the equation to standard form.

Slope: $\frac{45-31}{37-27} = \frac{14}{10} = 1.4$

Use point slope form: y - y₁ = m(x - x₁)

Plug in x = 27, y = 31, and m = 1.4

y - 31 = 1.4(x - 27)

y - 31 = 1.4x - 37.8

y = 1.4x - 6.8

Convert this into standard form: ax + by = c

y - 1.4x = - 6.8

5 0

1 year 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

2 years ago