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Rufina [12.5K]
3 years ago
15

I need a step-by-step process of how to solve this I know the answer but I need to know exactly how to solve these kinds of ques

tions so I know how to do it for a test:
A large ship is being towed by two tugs. The larger tug exerts a force which is 25 percent greater than the smaller tug, and at an angle of 30 degrees south of west.
Which direction must the smaller tug pull to ensure that the ship travels due west?
Mathematics
1 answer:
IgorLugansk [536]3 years ago
6 0
So what we know from that is that angle is 30 degrees south of west. so you go 30 degrees down from west. How can we do this? well <span>basically what you have to do is to break down the force into x and y component. The y component of large force and the y component of small force should cancel so that the boat doesn't go north and south and the x components of both the forces should add up so that it goes west only.</span>
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Find the missing angle
labwork [276]

Answer:

x=21

Step-by-step explanation:

69+90= 159

180-159=21

3 0
3 years ago
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How do you simplify each expression 3-5w+12+2w-3
Wittaler [7]
3-5w+12+2w-3
(3+12-3)=12
(-5w+2w)=-3w
-3w+12
5 0
3 years ago
A carpenter has less than 120 min. to spend painting furniture each day. Today, he has spent 30.5 min. painting a desk. Now he w
FinnZ [79.3K]

Answer:

Hence, a carpenter can paint a maximum of 7 chairs.

Hence, option A is correct.

Step-by-step explanation:

A carpenter has less than 120 min. to spend painting furniture each day.

Today, he has spent 30.5 min. painting a desk.

Now he will paint x chairs, each of which takes 12.5 min.

i.e. he will take a total of 12.5x minutes in painting 'x' chairs.

Also total time spent on painting furniture each day will be equal to time spent on painting a desk and all of the chairs.

Hence,

30.5+12.5x<120 (since the total time he spends on painting is less than 120 mins.)

⇒ 12.5 x<120-30.5

⇒ 12.5 x<89.5

⇒ x<7.16  ( on dividing both side of the inequality with 12.5)

Hence, a carpenter can paint a maximum of 7 chairs.


6 0
3 years ago
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Multiply and simplify. (x - 1)(x - 1)(x - 1)
Tanzania [10]
(sorry if it's too late and you've already figured it out, but here you go anyway)
The easiest way to do this is to start by FOILing then add.
So just start with (x-1)(x-1)
(x-1)(x-1)
Front: (x*x) = x^2
Outer: (x*-1) = -x
inner: (-1*x) = -x
Last: (-1*-1) = 1
Added: x^2 -2x +1
Now take that answer and do the same thing with (x-1). It's basically the same thing, just with an added thing you need to multiply.
(x-1)(x^2-2x+1)
(x*x^2) = x^3
(x*2x) = 2x^2
(x*1) = x
(-1*x^2) = -x^2
(-1*-2x) = 2x
(-1*1) = -1
Now add everything together:
x^3+2x^2+x-x^2+2x-1
The answer is:
x^3+x^2+3x-1
3 0
3 years ago
Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t&lt;3 if 3≤t&lt;5 if 5≤t&lt;[infinity],y(0)=4. y′+5y={0 if 0≤t&lt;311 i
rosijanka [135]

It looks like the ODE is

y'+5y=\begin{cases}0&\text{for }0\le t

with the initial condition of y(0)=4.

Rewrite the right side in terms of the unit step function,

u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t

In this case, we have

\begin{cases}0&\text{for }0\le t

The Laplace transform of the step function is easy to compute:

\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s

So, taking the Laplace transform of both sides of the ODE, we get

sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s

Solve for Y(s):

(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}

We can split the first term into partial fractions:

\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs

If s=0, then 1=5a\implies a=\frac15.

If s=-5, then 1=-5b\implies b=-\frac15.

\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}

\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}

Take the inverse transform of both sides, recalling that

Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)

where F(s) is the Laplace transform of the function f(t). We have

F(s)=\dfrac1s\implies f(t)=1

F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}

We then end up with

y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}

3 0
4 years ago
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