7 - (63/9) to the power of 2 divided by 7

An equation of the line tangent to y=sin x + 2cosx at (pi/2, 1) is?

Lina20 [59]

Answer:

B. 2x + y = pi + 1.

Step-by-step explanation:

y = sin x + 2 cos x

The slope of a line tangent is the derivative, so:

y' = cos x - 2 sin x

At (pi/2, 1) , substituting for x:-

cos pi/2 - 2 sin pi/2

= 0 - 2 * 1 = -2.

So the slope is - 2 and an equation can be found by using the point-slope formula:

y - y1 = m(x - x1) so:

y - 1 = -2(x - pi/2)

y = -2x + pi + 1

2x + y = pi + 1.

6 0

3 years ago

Solve the following differential equation: y" + y' = 8x^2

PtichkaEL [24]

Answer:

$y=y_p+y_h = \frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$

Step-by-step explanation:

Let's find a particular solution. We need a function of the form $y= ax^3+bx^2+cx+d$ such that

$y'= 3ax^2+2bx+c$ and

$y''= 6ax+2b$

$y'+y''= 3ax^2+2bx+c+6ax+2b = 3ax^2+x(2b+6a)+(c+2b) = 8x^2$

then, 3a= 8, 2b+6a =0 and c+2b = 0. With the first equation we obtain

a = 8/3 and replacing in the second equation 2b+6(8/3) = 2b + 16 = 0. Then, b = -8. Finally, c = -2(-8) = 16.

So, our particular solution is $y_p= \frac{8}{3}x^3-8x^2+16x$ .

Now, let's find the solution $y_p$ of the homogeneus equation $y''+y'=0$ with the method of constants coefficients. Let $y=e^{\lambda x}$

$y'=\lambda e^{\lambda x}$

$y''=\lambda^2e^{\lambda x}$

then $\lambda e^{\lambda x}+\lambda^2 e^{\lambda x} = 0$

$e^{\lambda x}(\lambda +\lambda^2)= 0$

$(\lambda +\lambda^2)= 0$

$\lambda (1+\lambda)= 0$

$\lambda =0$ and $\lambda)= -1$ .

So, $y_h = C_1 + C_2e^{-x}$ and the solution is

$y=y_p+y_h =\frac{8}{3}x^3-8x^2+16x+ C_1 + C_2e^{-x}$ .

5 0

3 years ago

.52 is 10 times as great as what number

WARRIOR [948]

Answer:

is it's right answer is 572

6 0

3 years ago

Which size would you see on the box for a new television whose screen measures 32 inches wide by 18 inches high?

zzz [600]

Answer:

37

Step-by-step explanation:

The calculation of size for a new television is shown below:-

With the help of Pythagoras theorem

$h^2 = b^2 + p^2$

where

h indicates hypotenuse which is the diagonal length

b indicates the base which is the length of the horizontal side

p indicates perpendicular which is the TV length of the vertical side.

$h^2 = \sqrt{b^2 + p^2}$

now, we will put the values into the above formula

$= \sqrt{32^2 + 18^2} \\\\ = \sqrt{1024 + 324} \\\\ = \sqrt{1348}$

Which gives result

= 36.7

or

= 37

5 0

3 years ago

Find parametric equations through point P=(2,2,7) in the direction of the vector v = (44, 14, -20)?

ycow [4]

9514 1404 393

Answer:

(x, y, z) = (2+44t, 2+14t, 7-20t)

Step-by-step explanation:

One way to write parametric equations for line L is ...

L = P + t·<em>v</em>

where P is the given point and <em>v</em> is the given direction vector. Using that form, we have ...

(x, y, z) = (2+44t, 2+14t, 7-20t)

__

If you like, you can remove a common factor of 2 from the coefficients of t.

(x, y, z) = (2+22t, 2+7t, 7-10t)

5 0

2 years ago