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2 years ago

Can you please help me I give you more points

Mathematics

1 answer:

never [62]2 years ago

8 0

Answer:

( 0 , -7 ) , (5 , -7 ) , (1 , -5 )

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18÷1,242 long division

vladimir2022 [97]

I hope this can help.

5 0

2 years ago

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

A Simple Random Sample reflects that ________ individual in the population has an ________ chance of being selected.

monitta

Answer:

any, equal

Step-by-step explanation:

A Simple Random Sample reflects that any individual in the population has an equal chance of being selected.

7 0

2 years ago

Car A, traveling at a constant 10 m/s, crosses the start line 3 seconds before Car B, who is traveling at a constant 15 m/s. At

Margaret [11]

The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m

Distance traveled by car A =x, (when the car B is at the same distance from the start line)
time of car A=t

x=10 m/st ⇒ x=10t (1)

Distance traveled by car B=x
time of car B=t-3

x=15(t-3) (2)

With the equations (1) and (2) we make a system of equations:

x=10t
x=15(t-3)

We solve this system of equations:

10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9

t-3=9-3=6

x=10 t=10 (9)=90

Answer: The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.

8 0

3 years ago

Something about evaluating numbers or something. Question is in the picture.

Law Incorporation [45]

The answer is 10 - 3√2

:) hope this helps!

3 0

2 years ago