3 Is (-5) x (-4) is negative or positive? Answer:

Two bicyclists are 7/8 of the way through a mile long tunnel when a train approaches the closer end at 40 mph. The riders take o

Anastaziya [24]

Answer:

the cyclists rode at 35 mph

Step-by-step explanation:

Assuming that the cyclists stopped, and accelerated instantaneously at the same speed than before but in opposite direction , then

distance= speed*time

since the cyclists and the train reaches the end of the tunnel at the same time and denoting L as the length of the tunnel :

time = distance covered by cyclists / speed of cyclists = distance covered by train / speed of the train

thus denoting v as the speed of the cyclists :

7/8*L / v = L / 40 mph

v = 7/8 * 40 mph = 35 mph

v= 35 mph

thus the cyclists rode at 35 mph

6 0

4 years ago

Use the method of "undetermined coefficients" to find a particular solution of the differential equation. (The solution found ma

Naddika [18.5K]

Answer:

The particular solution of the differential equation

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

Step-by-step explanation:

Given differential equation y''(x) − 10y'(x) + 61y(x) = −3796 cos(5x) + 185e6x

The differential operator form $(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Rules for finding particular integral in some special cases:-

let f(D)y = $e^{ax}$ then

the particular integral $\frac{1}{f(D)} (e^{ax} ) = \frac{1}{f(a)} (e^{ax} ) if f(a)$ ≠ 0

let f(D)y = cos (ax ) then

the particular integral $\frac{1}{f(D)} (cosax ) = \frac{1}{f(D^2)} (cosax ) =\frac{cosax}{f(-a^2)}$ f(-a^2) ≠ 0

Given problem

$(D^{2} -10D+61)y(x) = −3796 cos(5x) + 185e^{6x}$

Particular integral:-

$P.I = \frac{1}{f(D)}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + 185e^{6x})$

$P.I = \frac{1}{D^2-10D+61}( −3796 cos(5x) + \frac{1}{D^2-10D+61}185e^{6x})$

P.I = $I_{1} +I_{2}$

we will apply above two conditions, we get

$I_{1}$ =

$\frac{1}{D^2-10D+61}( −3796 cos(5x) = \frac{1}{(-25)-10D+61}( −3796 cos(5x) ( since D^2 = - 5^2)$ $= \frac{1}{(36-10D}( −3796 cos(5x) \\= \frac{1}{(36-10D}X\frac{36+10D}{36+10D} ( −3796 cos(5x)$

on simplification we get

= $\frac{1}{(36^2-(10D)^2}36+10D( −3796 cos(5x)$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{1296-100(-25)}$

= $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$

$I_{2}$ =

$\frac{1}{D^2-10D+61}185e^{6x}) = \frac{1}{6^2-10(6)+61}185e^{6x})$

$\frac{1}{37}185e^{6x})$

Now particular solution

P.I = $I_{1} +I_{2}$

P.I = $\frac{-1,36,656cos5x+1,89,800 sin5x}{-1204}$ + $\frac{1}{37}185e^{6x})$

8 0

3 years ago

Can any one help me plzzz and thanks

GarryVolchara [31]

Answer:

----

ac

Step-by-step explanation:

8 0

4 years ago

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Can i get a OHHHHHH YEAHHHH

Darina [25.2K]

Answer:

AW YEAHHHHHHHHHHHHHHHHHHHHHHHHHHH

Step-by-step explanation:

8 0

3 years ago

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***Will mark all right answers brainliest*** A certain type of bacteria is being grown on a Petri dish in the school’s biology l

alex41 [277]

Answer:

On day 0 (starting day), the percentage of petri dish occupied by bacteria was 2.44%

Step-by-step explanation:

Rate of growth = 2 (i.e. doubles every day)

Petri dish was filled to 100% on day 12.

Let

P(0) = percentage of Petri dish occupied on day 0, then

equation of percentage a function of time in x days

P(x) = P(0)*r^x ......................(1)

where

100% = P(12) = p(0) * 2^12 = 4096 P(0)

=>

P(0) = 100% / 4096 = 0.0244%

Next, to find percentage on February 14 (Valentine's day!)

Day 0 is February 9, so February 14 is the fifth day, so x=5.

Substitute x=5 in equation (1) above,

P(x) = P(0)*r^x

P(5) = P(0)*2^5

P(5) = 0.0244*2^5 = 0.0244*32 = 0.781%

Ans. the 0.781% of the petri dish was filled with bacteria after 5 days on February 14th.

4 0

4 years ago

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