(85pts)

Mathematics

1 answer:

RUDIKE [14]2 years ago

8 0

Step-by-step explanation:

1. The first step has 1 block. The second step has 2 blocks. The third step has 3 blocks. So the nth step needs n blocks.

2. A staircase with n steps has 1 + 2 + 3 + ... + n steps. This is an arithmetic sequence where the first term is 1 and the last term is n. The sum of the first n terms is:

y = n/2 (1 + n)

3. Solve for n.

2y = n (1 + n)

2y = n² + n

0 = n² + n − 2y

Using quadratic formula:

n = [ -1 ± √(1 − 4(1)(-2y)) ] / 2

n = [ -1 + √(1 + 8y) ] / 2

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What number goes in the box below to make the statement true?

Schach [20]

Answer:

5000

Step-by-step explanation:

4 0

3 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$