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Solve for x assume that lines appear tangent are tangent

Anuta_ua [19.1K]

Answer:

Step-by-step explanation:

? = 1/2(230)

A 115°

8 0

2 years ago

When dealing with word problems and linear equations, what does the y-intercept represent and how is the slope defined?

katovenus [111]

Answer:

C

Step-by-step explanation:

y-intercept represents the starting value, slope is the rate of change

slope is rise/run

6 0

2 years ago

Read 2 more answers

Graph by using the slope and the y intercept.

Mashutka [201]

1. a) equation of the line :

$y = \dfrac{2}{5}x - 7$

y - intercept = -7

so, it will pass through point (0, -7)

and if we plug the value of x as 5, we get

$y = (\dfrac{2}{5} \times 5) - 7$

$y = 2 - 7$

$= - 5$

so, it will pass through point (5, -5) too

now, just plot the points (0 , -7) and (5 , -5) and join them.

2. b) equation of line is :

$y = - 3x + 5$

here, y - intercept = 5

so the line passes through point (0 , 5)

now, Plugging the value of x = 1 we get :

$y =( - 3 \times 1) + 5$

$y = - 3 + 5$

$y = 2$

so, the given line passes through point (1 , 2)

plotting the points, we can get our required line.

6 0

2 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

Which expression is equivalent to –2(5x – 0.75)?

zzz [600]

-10x + 1.5 because you get rid of the parenthesis so -2 times 5 is -10 ( -10x ) and then -2 times -0.75 equals positive 1.5 due to two negatives canceling out to positive hope this helped!

4 0

2 years ago

Read 2 more answers