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Dominik [7]
3 years ago
11

Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio

n. Do not attempt to find the solution. (Enter your answer using interval notation.) (x − 2)y'' + y' + (x − 2)(tan x)y = 0, y(3) = 1, y'(3) = 4

Mathematics
1 answer:
natali 33 [55]3 years ago
5 0

Answer:

answer in image. Thanks

Step-by-step explanation:

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X=4, y= -7, z=9 solve: -2x-3y+z
vesna_86 [32]

Answer:

22

Step-by-step explanation:

To solve this we can plug the values in.

-2x-3y+z

-2(4)-3(-7)+(9)

Simplify.

-8+21+9

Add.

22

8 0
2 years ago
Draw a number line to<br><br> represent the inequality.<br><br> y s 23
Mkey [24]

Answer:

The number lines are shown below.

Step-by-step explanation:

In the given problem the sign of inequality is missing.

We know, that there are 4 signs of inequality ">", "<", "≥" and "≤".

The possible inequalityes are

y>23

y

y\geq 23

y\leq 23

In y>23, all points on the right side of 23 are included in the solution set.

In y, all points on the left side of 23 are included in the solution set.

In y\geq 23, 23 and all points on the right side of 23 are included in the solution set.

In y\leq 23, 23 all points on the left side of 23 are included in the solution set.

4 0
3 years ago
Order the side lengths from least to greatest
Liula [17]

Answer:

BC < CE < BE < ED < BD

Step-by-step explanation:

In the triangle BCE,

m∠BEC + m∠BCE + m∠CBE = 180°

m∠BEC + 81° + 54° = 180°

m∠BEC = 180 -  135

m∠BEC = 45°

Order of the angles from least to greatest,

m∠BEC < m∠CBE > mBCE

Sides opposite to these sides will be in the same ratio,

BC < CE < BE ----------(1)

Now in ΔBED,

m∠BEC + m∠BED = 180°

m∠BED = 180 - 45

             = 135°

Now, m∠BDE + m∠BED + DBE = 180°

11° + 135°+ m∠DBE = 180°

m∠DBE = 180 - 146

             = 34°

Order of the angles from least to greatest will be,

∠BDE < ∠DBE < ∠BED

Sides opposite to these angles will be in the same order.

BE < ED < BD ----------(2)

From relation (1) and (2),

BC < CE < BE < ED < BD

7 0
3 years ago
There are 12 students in a classroom, including the triplets joey, chloe, and zoe. If 3 of the 12 are randomly selcted to give s
BARSIC [14]
The probability of Joey being first is 1/12. If that happens the probability of Chloe going next is 1/11. Then the probability of zoe being picked next would be 1/10. Then if you multiply all the fractions I gave you you would get 1/1320 which is A. So A. Is your answer!! Hope this helps you out!!
8 0
3 years ago
Read 2 more answers
. Adult tickets to a play cost $22. Tickets for children cost $15. Tickets for a group of
Natalija [7]
2 x 15 = 30, 9 x 22 = 198, 198 + 30 = 228.
So the answer is A 2 children, 9 adults.

Hope I helped, Ms. Weasley
If you feel that I deserve it I would love to be marked Brainliest! 

8 0
3 years ago
Read 2 more answers
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