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3 years ago

Last 2 tutor find it wrong pls help me

Mathematics

1 answer:

Mrac [35]3 years ago

4 0

Answer:

g(x) = 4x² -x +64

Step-by-step explanation:

The integral of f(x) will be continuous. This means the value must be the same for x = 4- as for x = 4+.

For some constant c1, the integral in the region x ≤ 4 is ...

∫f(x)dx = (4x² -x +c1) +C = g(x) +C

The integral in the region x ≥ 4 is given as ...

∫f(x)dx = 31x +C

Evaluated at x=4, we must get the same value from either expression:

4(4²) -4 +c1 +C = 31(4) +C

60 +c1 = 124

c1 = 64

In order to make the integral continuous at x=4, we must have ...

g(x) = 4x² -x +64

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Math help please help thanks

ser-zykov [4K]

Answer:

Step-by-step explanation:

Draw a line from the bottom of the 4 foot line straight across (horizontally) until it hits the vertical line on your left.

The two lines (the one you drew and the left vertical line) meet at right angles.

The upper figure is a rectangle.

Area = L * W

L = 5

W = 4

Area = 5 * 4

Area = 20

Now the bottom rectangle can be found the same way.

Area = L * W

L = 14

W = 3

Area = 14 * 3

Area = 42

The total area = 42 + 20

Total area = 62.

The comment was correct.

5 0

3 years ago

A small rocket is fired from a launch pad 10 m above the ground with an initial velocity left angle 250 comma 450 comma 500 righ

jonny [76]

Let $\vec r(t),\vec v(t),\vec a(t)$ denote the rocket's position, velocity, and acceleration vectors at time $t$ .

We're given its initial position

$\vec r(0)=\langle0,0,10\rangle\,\mathrm m$

and velocity

$\vec v(0)=\langle250,450,500\rangle\dfrac{\rm m}{\rm s}$

Immediately after launch, the rocket is subject to gravity, so its acceleration is

$\vec a(t)=\langle0,2.5,-g\rangle\dfrac{\rm m}{\mathrm s^2}$

where $g=9.8\frac{\rm m}{\mathrm s^2}$ .

a. We can obtain the velocity and position vectors by respectively integrating the acceleration and velocity functions. By the fundamental theorem of calculus,

$\vec v(t)=\left(\vec v(0)+\displaystyle\int_0^t\vec a(u)\,\mathrm du\right)\dfrac{\rm m}{\rm s}$

$\vec v(t)=\left(\langle250,450,500\rangle+\langle0,2.5u,-gu\rangle\bigg|_0^t\right)\dfrac{\rm m}{\rm s}$

(the integral of 0 is a constant, but it ultimately doesn't matter in this case)

$\boxed{\vec v(t)=\langle250,450+2.5t,500-gt\rangle\dfrac{\rm m}{\rm s}}$

and

$\vec r(t)=\left(\vec r(0)+\displaystyle\int_0^t\vec v(u)\,\mathrm du\right)\,\rm m$

$\vec r(t)=\left(\langle0,0,10\rangle+\left\langle250u,450u+1.25u^2,500u-\dfrac g2u^2\right\rangle\bigg|_0^t\right)\,\rm m$

$\boxed{\vec r(t)=\left\langle250t,450t+1.25t^2,10+500t-\dfrac g2t^2\right\rangle\,\rm m}$

b. The rocket stays in the air for as long as it takes until $z=0$ , where $z$ is the $z$ -component of the position vector.

$10+500t-\dfrac g2t^2=0\implies t\approx102\,\rm s$

The range of the rocket is the distance between the rocket's final position and the origin (0, 0, 0):

$\boxed{\|\vec r(102\,\mathrm s)\|\approx64,233\,\rm m}$

c. The rocket reaches its maximum height when its vertical velocity (the $z$ -component) is 0, at which point we have

$-\left(500\dfrac{\rm m}{\rm s}\right)^2=-2g(z_{\rm max}-10\,\mathrm m)$

$\implies\boxed{z_{\rm max}=125,010\,\rm m}$

7 0

3 years ago

Which list puts the following things in correct order from smallest to largest?

slega [8]

Answer:

A. "hope this helps"

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

if any one can help me that would be awesome 1. − 8 = 5 b − 3 b 2. k − 8 + 6 k = 20 3. 723. 4. 6 ( 3 + 3 x ) = 72 364. 5 ( m + 4

Morgarella [4.7K]

Answer:

1)b=-4

2)k=4

3)can u explain what #3 and 4 say

4)?

5)0/ no solution

Step-by-step explanation:

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3 years ago

M divided by 19 = -2<br><br> (Find m)

seropon [69]

M=38 I hope it helps
Can u give me the brainliest ^^ plz ☺️☺️

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3 years ago