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3 years ago

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There are (22)4 ⋅ 20 cows in a meadow. What is the total number of cows in the meadow?

1 answer:

Archy [21]3 years ago

6 0

The answer will be i think c

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2. A citrus grower ships grapefruit in boxes of

amid [387]

I think it’s
20,400*10=204,000

8 0

3 years ago

A wall is built of large blocks that are 1 foot by 1 foot by 1 foot,and each weigh 100 pounds. The volume of the wall is 240 cub

LenaWriter [7]

If 1 cubic feet=100 pounds then, 240 cubic feet=x pounds

1=100
240=x

cross multiply; x=24000 so x=24000

5 0

4 years ago

The oak tree at school was 34 feet high it grew 18 feet taller how tall is the oak tree now

Alex_Xolod [135]

34 feet + 18 feet = 52 feet
The oak tree is 52 feet tall.

4 0

3 years ago

Read 2 more answers

ANSWER NEEDED NOW!!!

Sunny_sXe [5.5K]

Answer is A i hope it helps

7 0

4 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:

<em>y''</em> + 4<em>y'</em> + 4<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 4 = (<em>r</em> + 2)² = 0

with a double root at <em>r</em> = -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 <em>ansatz</em>

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

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 6<em>A</em> = 12 and 2<em>B</em> = 0, or <em>A</em> = 2 and <em>B</em> = 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 4<em>C</em> = -8 and 4<em>C</em> + 4<em>D</em> = -12, or <em>C</em> = -2 and <em>D</em> = -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 <em>y</em> (0) = -2 and <em>y'</em> (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