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2 years ago

HELPPPPPPPP PLEASEEEEEEE!!!!!!!

Mathematics

1 answer:

Alja [10]2 years ago

5 0

Answer:

false

Step-by-step explanation:

it's false negative

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Jake can carry 6 1/4 pounds of wood in from the barn. His father can carry 1 5/7 times as much as jake. How many pounds can jake

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Jake Pauls Dad Can Carry 10.7 Pounds

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3 years ago

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The graph of the equations was plotted using geogebra graphing and attached.

Let x represent the hours weightlifting and y represent the hours doing cardio exercises.

Since he spend a maximum of 20 hours, hence:

x + y ≤ 20 (1)

Also, he spends at least 8 of those hours weightlifting, hence:

x ≥ 8 (2)

He wants to spend no more than 15 hours doing cardio exercises, this is:

y ≤ 15 (3)

The graph of the equations was plotted using geogebra graphing and attached.

Find out more at: brainly.com/question/17178834

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3 years ago

Sierra has 3 feet of ribbon to make bows for friends.She uses 1/5 of a foot of ribbon for each bow. She think she can make more

lubasha [3.4K]

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2 years ago

Read 2 more answers

Tickets to a concert cost $50 for each orchestra seat and $40 for each

Vsevolod [243]

Answer:

H. the number of orchestra seat is 900

Step-by-step explanation:

Step one:

let the number of orchestra seat be x

and balcony seat be y

cost of orchestra= $50 each

cost of balcony =$40 each

total tickets= 1500

x+y= 1500----------1

amount earned= $69000

50x+40y=69000--------2

The system of equation for the situation is

x+y= 1500----------1

50x+40y=69000--------2

from 1, x=1500-y

put this in equation 2

50(1500-y)+40y=69000

75000-50y+40y=69000

-10y=69000-75000

-10y=-6000

divide both sides by -10

y=-6000/-10

y=600

put y= 600 in equation 1

x+600= 1500

x=1500-600

x=900

6 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago