HELP PLEASEEEE<br> what is 3x(-1)-1 ?

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):

$y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1 \right)} \, dx$

$y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx$

$y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C$

$C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0$

$C = 0$

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

5 0

3 years ago

You lose 2 points every time you forget to write your name on a test. You have forgotten to write your name 4 times. What intege

Fed [463]

Answer:

The answer is -8

I hope this helps you :D

5 0

3 years ago

Read 2 more answers

Help if your smart plzzzzzzz <br> ( will give 35 pts and brainliest)

Temka [501]

Answer:

a) alternate interior angles theorem

b) OXP ≅ XOL

c) XO ≅ OX

d) reflexive property (i'm not sure about this one)

e) ΔXOP ≅ ΔOXL

f) cpctc

make sure to double check the fourth one

4 0

2 years ago

Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the

kodGreya [7K]

Answer:

p=0.7103 (4-game series)
p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters n=4, p=0.6:

$P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\ i=4,5,6,7$

#We find probabilities for the different values of i:

$P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659$