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2 years ago

Need help with area for solving this

Mathematics

2 answers:

Dmitriy789 [7]2 years ago

8 0

Answer:

Step-by-step explanation:

First find the area of total rectangle which equals 20 * 12 = 240

Second find the area of triangle which equals 0.5 * 4 * 6 = 12

The required area = total rectangle - triangle = 240 - 12 = 228

Bumek [7]2 years ago

8 0

Answer:

NEEEEEGARRRRR

Step-by-step explanation:

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Is 4/8 greater or less than 4/6

IrinaK [193]

Find the lowest common multiple=
8= 8,16,24,32,40,48
6=6,12,18,24,30,36
the lowest common multiple is 24 so then
4/8>>>>?/24 you times 8 by 3 to get 24 so you times 4 by 3 which equals 12 so it would be 12/24 (or a half)
then
4/6>>>>?/24 you times 6 by 4 to get 24 so you times 4 by 4 which equals 16 so it would be 16/24

12/24(4/8) is greater or less than 16/24(4/6)
it is less than. :)

5 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

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

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

7 0

2 years ago

Given two sides with lengths 34 and 14, which of these lengths would NOT work for the third side of a triangle? 18, 34, 44, 22,

Bas_tet [7]

Two sides are 34 and 14.

The third side can't be longer than (34 + 14) = 48, because the 34 and the 14
together could not reach from one end of it to the other.

The third side also can't be shorter than 20, because the 20 and the 14 together
could not reach from one end of the 34 to the other.

So 18 doesn't work. (18 + 14) = 32 ... they couldn't cover the 34 .

7 0

2 years ago

Ok remember these things im saying before you do the answer A=1 B=3 C=2 D=4 E=5 What is A-6+D-C-B=?

Marat540 [252]

Answer:

the answer is 0

5 0

2 years ago

Read 2 more answers

Distance traveled in 3 hours at 35km/h

Ghella [55]

I hope this helps you

Distance = speed.time

Distance=35.3

Distance=105