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katen-ka-za [31]
3 years ago
13

Determine whether the given function is a solution to the given differential equation.

Mathematics
1 answer:
Anarel [89]3 years ago
5 0

Answer:

It is only 1 and 2 that is the solution to the differential equation

Step-by-step explanation:

1. x''=- 2 Cos t+ 3 sin t and x=2 cos t -3 sin t the addition give 0

2. d^2y/dx^2=- Sin x +2 and y =Sin x +x^2 the addition =x^2+2

but

3. dx/dt= -2 Sin 2t and tx =t Cos 2t =t( 2Cos t - 2 Sin t) the addition it can't equal Sin 2t

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What is the input of 3x-7 when it equals -1
AveGali [126]

Answer:

21

Step-by-step explanation:

3x-7=-21

-21x-1=21

hope this helps, brainliest is always appreciated :3

8 0
2 years ago
Can someone please help me
Misha Larkins [42]
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6 0
3 years ago
Read 2 more answers
Harold has 4 classes each morning. Each class is 1 hour long, and there are 10 minutes between classes. The first class is at 8
Viktor [21]

The fourth class ends at 12:30 pm

<em><u>Solution:</u></em>

Given that Harold has 4 classes each morning

Each class is 1 hour long, and there are 10 minutes between classes

The first class is at 8 A.M

<em><u>To find: Time at which fourth class ends</u></em>

Since each is 1 hour long and 10 minutes gap between classes

First class = 8 A.M to 9 A.M

Second class = 9:10 A.M to 10 : 10 AM

Third class = 10 : 20 AM to 11 : 20 AM

Fourth class = 11 : 30 AM to 12 : 30 PM

Thus the fourth class ends at 12:30 pm

4 0
3 years ago
Use Gauss-Jordan elimination to solve the following linear system.
marysya [2.9K]
Just put the coefients in to a matrix

1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14

\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]
mulstiply 2nd row by -1 and add to 3rd
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]
divde last row by 2
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 2rd row by 6 and add to top one
\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 1st row by -1 and add to 2nd
\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]
divide 2nd row by -3
\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]
mulstiply 2nd row by -1 and add to 1st row
\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]
divide 1st row by -3
\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]

rerange
\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer
7 0
3 years ago
What is the constant proportionality of 3y = 2x
Ivenika [448]

Answer:

1.5

Step-by-step explanation:

3 0
3 years ago
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