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ivanzaharov [21]
3 years ago
10

In order to keep a boat from drifting away, the boat is tied to a pier with a 60-foot rope. When the boat drifts as far from the

pier as the rope will allow, the rope forms a 15° angle relative to the horizontal water line.
To the nearest foot, find the maximum horizontal distance, d, that the tied-up boat can drift from the pier.

16 feet
37 feet
58 feet
62 feet
Mathematics
1 answer:
Rus_ich [418]3 years ago
6 0
The answer is C) hope this helped!
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Sati [7]

Answer:

6x+20 or 2(3x+10)

Step-by-step explanation:

Six friends went to a basketball game and four people paid $5 extra for a t-shirt.  To find the total cost, multiply 6 by x and 4 by 5 this add them together:

6*x+4*5

6x+20, If needed you can factor this expression to:

2(3x+10)

7 0
3 years ago
12+ brainliest pppllleeeaaassee
Veronika [31]
The answer would be wx(yz) because the associative property says that you can change where the parenthesis are while keeping the same equation.
The equation is w(xy)z, but moving the parenthesis some where else will not change it, such as wx(yz).
3 0
3 years ago
Brainiest will be given . Please help me
mixas84 [53]

Answer:

If the shape of end face has n sides

Then, in total, there are n+2 faces

Using that formula

Faces in 20-sided polygon = n+2 = 20+2 = 22

Step-by-step explanation:

Pls vote as brainliest

6 0
3 years ago
A certain country has $10 billion in paper currency in circulation, and each day $50 million comes into the country's banks. The
mash [69]

Answer:

\bf x(t)=10(1-e^{-0.005*t})

Step-by-step explanation:

The differential equation

\bf \displaystyle\frac{dx}{dt}=0.005(10-x)

can be solved by separation of variables. Write the equation as

\bf \displaystyle\frac{dx}{10-x}=0.005dt

Integrate on both sides

\bf \int\displaystyle\frac{dx}{10-x}=\int0.005dt\Rightarrow -ln(10-x)=0.005t+C\Rightarrow\\\\\Rightarrow ln(10-x)^{-1}=0.005t+C\Rightarrow (10-x)^{-1}=e^{0.005t}e^C

where C is a constant.

\bf e^C is also a constant and we will keep calling it C, (there is no reason to change the letter). We have then

\bf (10-x)^{-1}=Ce^{0.005t}\Rightarrow \displaystyle\frac{1}{10-x}=Ce^{0.005t}\Rightarrow 10-x=\displaystyle\frac{1}{Ce^{0.005t}}\Rightarrow\\\\\Rightarrow x(t)=10-(1/C)e{-0.005t}

(1/C) is a constant, and for the same reason we will keep calling it C. So the general solution is

\bf x(t)=10-Ce^{-0.005t}

Now, we use the initial condition x(0)=0

\bf x(0)=10-Ce^{-0.005*0}=0\Rightarrow C=10

and the particular solution is

\bf x(t)=10-10e^{-0.005*t}=10(1-e^{-0.005*t})\\\\\boxed{x(t)=10(1-e^{-0.005*t})}

7 0
3 years ago
Help needed please!!!!!!!!
Alisiya [41]
I think the answer is B. Hope I helped
3 0
3 years ago
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