Ask question

Ask question

All categories

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

1 answer:

natali 33 [55]3 years ago

5 0

Answer:

answer in image. Thanks

Step-by-step explanation:

You might be interested in

Round 7.21 to the nearest whole number.

gtnhenbr [62]

It’s 7...your answer is 7

8 0

3 years ago

Read 2 more answers

The object above is symmetrical through Y = 17 inches Z = 18 inches, and H = 11 inches, what is the area of the object ?

oksian1 [2.3K]

Answer:

let me try

a= 1/2(l×w)

=1/2*17*18

=1/2(306)

A=153

8 0

3 years ago

Given the preimage and image,find the dilation scale factor.

Oliga [24]

Answer:

The dilation scale factor is $\frac{1}{2}$ .

Step-by-step explanation:

The image is the dilated form of its preimage if and only if the following conditions are observed:

1) $K' = \alpha_{1} \cdot K$

2) $T' = \alpha_{2} \cdot T$

3) $P' = \alpha_{3} \cdot P$

4) $J' = \alpha_{4} \cdot J$

5) $\alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}$

If we know that $K = (2, 0)$ , $K' = (1, 0)$ , $T = (3, 0)$ , $T' = (1.5,0)$ , $P = (1, 5)$ , $P' = (0.5, 2.5)$ , $J = (0, 3)$ and $J' = (0, 1.5)$ , then the coefficients are, respectively:

$\alpha_{1} = \frac{1}{2}$ , $\alpha_{2} = \frac{1}{2}$ , $\alpha_{3} = \frac{1}{2}$ , $\alpha_{4} = \frac{1}{2}$

As $\alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}$ , we conclude that the dilation scale factor applied in the preimage is equal to $\frac{1}{2}$ .

7 0

3 years ago

Mrs. Pender shows her students 2 sticks measuring 6 inches and 2 sticks measuring 8 inches. She asks the students to arrange the

Llana [10]

Answer:

Reggies makes sence

Step-by-step explanation:

5 0

3 years ago

Solve the system of linear equations. separate the x- and y- values with a coma. -8x=72-16y

serious [3.7K]

Let's use the "elimination by addition/subtraction" method to solve this system for x and y.

<span> -8x=72-16y
4x=-92+16y <= Mult. all 3 terms by 2: 8x = - 184 + 32y

Now combine the following, column by column:

</span><span> -8x= 72-16y
</span>8x = - 184 + 32y
-----------------------
0 = -112 + 16y. Solving for y, 16y = 112, and y = 7.

Now subst. 7 for y in either of the original equations and calculate x:

<span>4x=-92+16(7) => 4x = -92 + 112 = 20. Then x = 5.

Solution is 5, -7 (or, better written) (5, - 7).</span>

8 0

3 years ago