A rocket was launched from the ground and reached its maximum height of 36 feet at

Martina finished her chemistry assignment in 2/5 hours. Then she completed her math assignment in 1/4 hrs. What was the total am

JulsSmile [24]

$\text {Total time spent :}$

$\dfrac{2}{5} + \dfrac{1}{4}$

$= \dfrac{2 \times 4}{5 \times 4} + \dfrac{1 \times 5}{4 \times 5}$

$= \dfrac{8}{20} + \dfrac{5}{20}$

$= \dfrac{13}{20} \text { hour}$

4 0

3 years ago

Please. You can help me with this problem. Thanks.

erma4kov [3.2K]

Answer:

A

Step-by-step explanation:

3 0

4 years ago

Read 2 more answers

Please help me with this homework

Vera_Pavlovna [14]

Answer:

linear

non-linear

Step-by-step explanation:

8 0

3 years ago

In this problem, you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+4y=12te^(−2t)−(8t+12) with

Zarrin [17]

First check the characteristic solution:

y'' + 4y' + 4y = 0

has characteristic equation

r ² + 4r + 4 = (r + 2)² = 0

with a double root at r = -2, so the characteristic solution is

$y_c = C_1e^{-2t} + C_2te^{-2t}$

For the particular solution corresponding to $12te^{-2t}$ , we might first try the ansatz

$y_p = (At+B)e^{-2t}$

but $e^{-2t}$ and $te^{-2t}$ are already accounted for in the characteristic solution. So we instead use

$y_p = (At^3+Bt^2)e^{-2t}$

which has derivatives

${y_p}' = (-2At^3+(3A-2B)t^2+2Bt)e^{-2t}$

${y_p}'' = (4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t}$

Substituting these into the left side of the ODE gives

$(4At^3+(-12A+4B)t^2+(6A-8B)t+2B)e^{-2t} + 4(-2At^3+(3A-2B)t^2+2Bt)e^{-2t} + 4(At^3+Bt^2)e^{-2t} \\\\ = (6At+2B)e^{-2t} = 12te^{-2t}$

so that 6A = 12 and 2B = 0, or A = 2 and B = 0.

For the second solution corresponding to $-8t-12$ , we use

$y_p = Ct + D$

with derivative

${y_p}' = C$

${y_p}'' = 0$

Substituting these gives

$4C + 4(Ct+D) = 4Ct + 4C + 4D = -8t-12$

so that 4C = -8 and 4C + 4D = -12, or C = -2 and D = -1.

Then the general solution to the ODE is

$y = C_1e^{-2t} + C_2te^{-2t} + 2t^3e^{-2t} - 2t - 1$

Given the initial conditions y (0) = -2 and y' (0) = 1, we have

$-2 = C_1 - 1 \implies C_1 = -1$

$1 = -2C_1 + C_2 - 2 \implies C_2 = 1$

and so the particular solution satisfying these conditions is

$y = -e^{-2t} + te^{-2t} + 2t^3e^{-2t} - 2t - 1$

or

$\boxed{y = (2t^3+t-1)e^{-2t} - 2t - 1}$

7 0

3 years ago

In 1998 the U.S. Hispanic high school dropout rate was 29.5% and in 2008 it had dropped to 18.3%.

Bogdan [553]

A. The drop-out rate can be estimated using the equation,

d = my + b

where d is the drop-out rate, m is the slope of the line, y is the number of years from 1998 and b be the initial value. For 2008,

18.3 = m(2008 - 1998) + 38.46

The value of m from the equation is -1.12. The equation becomes,

d = -1.12y + 38.46

b. The vertical intercept of the model is when d is equal to zero.

0 = -1.12y + 38.46

The value of y from the equation is 34.34.
(34.34, 0)

c. n = 1990 + 34.34 = 2024.34.

The vertical intercept of this problem tells me that by the year 2024, the drop-out rate would be 0%.

5 0

3 years ago