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2 years ago

The force in Newtons acting on a particle located x meters from the origin is given by the function F(x)=12x2+7. What is the wor

k done in Joules when moving the particle from x=2 to x=5?

Mathematics

1 answer:

Shkiper50 [21]2 years ago

5 0

$W = \displaystyle \int^{5}_{2} F(x) ~ dx\\\\\\= \displaystyle \int^{5}_{2} \left(12x^2 +7\right) ~dx\\\\\\= 12\displaystyle \int^{5}_{2} x^2 ~dx + \displaystyle \int^{5}_{2} 7~ dx\\\\\\=12 \left[\dfrac{x^3}3 \right]^{5}_{2} + 7\left[x\right]^{5}_{2}\\\\\\=4(5^3-2^3) + 7(5-2) = 468 + 21 = 489~ J$

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Captain Emily has a ship, the H.M.S Crimson Lynx. The ship is five furlongs from the dread pirate Umaima and her merciless band

LenaWriter [7]

Answer:

The probability that the pirate misses the captain's ship but the captain hits = 0.514

Step-by-step explanation:

Let A be the event that the captain hits the pirate ship

The probability of the captain hitting the pirate ship, P(A) = 3/5

Let B be the event that the pirate hits the captain's ship

The probability of the pirate hitting the captain's ship P(B) = 1/7

The probability of the pirate missing the captain's ship, P'(B) = 1 - P(B)

P'(B) = 1 - 1/7 = 6/7

The probability that the pirate misses the captain's ship but the captain hits = P(A) * P(B) = 3/5 * 6/7

= 0.514

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3 years ago

What is the answer I need help ASAP

ira [324]

Answer:

C) 4

Step-by-step explanation:

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2 years ago

What are the vertical asymptotes of the function f(x) =5x+5/x2 + x-2

katen-ka-za [31]

let's recall that the vertical asymptotes for a rational expression occur when the denominator is at 0, so let's zero out this one and check.

$\bf \cfrac{5x+5}{x^2+x-2}\qquad \stackrel{\textit{zeroing out the denominator}~\hfill }{x^2+x-2=0\implies (x+2)(x-1)=0}\implies \stackrel{\textit{vertical asymptotes}}{ \begin{cases} x=-2\\ x=1 \end{cases}}$

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2 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

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