Ask question

Ask question

All categories

2 years ago

8

There are two cables coming off the top of a building from the same point. Each cable is secured to the ground. The longest cabl

e is 200 ft. It is attached at point M and creates a 32° angle of elevation with the ground. The other cable is attached at point N. If the angle formed between the two cables at the top of the building is 25°, what is the distance between points M and N?
Measure of Ð N =
°

2 answers:

Bas_tet [7]2 years ago

6 0

Call the point of attachment at the top of the building point T. (Angle T = 25°) The angle N = 180° -32° -25° = 123°. The Law of Sines tells you the relationship ...

... TM/sin(N) = MN/sin(T)

... MN = TM·sin(T)/sin(N) = (200 ft)·sin(25°)/sin(123°)

... MN ≈ 100.8 ft

USPshnik [31]2 years ago

3 0

Answer:

N= 123

x= 100.8

Step-by-step explanation:

I just did it

You might be interested in

If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?

kiruha [24]

Answer:

$[1,\infty)$

Step-by-step explanation:

$b(x)=\sqrt{x-4}$

$a(x)=3x+1$

Since we want to know the domain of $(b \circ a)(x)$ , let's first consider the domain of the inside function, that is, that of $a(x)=3x+1$ . Every polynomial function has domain all real numbers.

So we can plug anything for function $a$ and get a number back.

Now the other function is going to be worrisome because it has a square root. You cannot take square root of negative numbers if you are only considering real numbers which that is the case with most texts.

Let's find $(b \circ a)(x)$ and simplify now.

$(b \circ a)(x)$

$b(a(x))$

$b(3x+1)$

$\sqrt{(3x+1)-4}$

$\sqrt{3x+1-4}$

$\sqrt{3x-3}$

Now again we can only square root positive or zero numbers so we want $3x-3 \ge 0$ .

Let's solve this to find the domain of $(b \circ a)(x)$ .

$3x-3 \ge 0$

Add 3 on both sides:

$3x \ge 3$

Divide both sides by 3:

$x \ge 1$

So we want $x$ to be a number greater than or equal to 1.

The option that says this is $[1,\infty)$

-------------------------------

Give an example why option A fails:

A number in the given set is -2.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

So $a(-2)=3(-2)+1=-6+1=-5$ and $b(-5)=\sqrt{-5-4}=\sqrt{-9} \text{ which is not real}$ .

Give an example why option B fails:

A number in the given set is 0.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

So $a(0)=3(0)+1=0+1=1$ and $b(1)=\sqrt{1-4}=\sqrt{-3} \text{ which is not real}$ .

Give an example why option D fails:

While all the numbers in set D work, there are more numbers outside that range of numbers that also work.

A number not in the given set that works is 3.

$a(x)=3x+1$

$b(x)=\sqrt{x-4}$

So $a(3)=3(3)+1=9+1=10$ and $b(1)=\sqrt{10-4}=\sqrt{6} \text{ which is real}$ .

4 0

2 years ago

Read 2 more answers

A common assumption in modeling drug assimilation is that the blood volume in a person is a single compartment that behaves like

mixas84 [53]

Answer:

a) $\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) $\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) the steady state mass of the drug is 2000 mg

d) t ≅ 153.51 minutes

Step-by-step explanation:

From the given information;

At time t= 0

an intravenous line is inserted into a vein (into the tank) that carries a drug solution with a concentration of 500

The inflow rate is 0.06 L/min.

Assume the drug is quickly mixed thoroughly in the blood and that the volume of blood remains constant.

The objective of the question is to calculate the following :

a) Write an initial value problem that models the mass of the drug in the blood for t ≥ 0.

From above information given :

$Rate _{(in)}= 500 \ mg/L \times 0.06 \ L/min = 30 mg/min$

$Rate _{(out)}=\dfrac{x}{4} \ mg/L \times 0.06 \ L/min = 0.015x \ mg/min$

Therefore;

$\dfrac{dx}{dt} = Rate_{(in)} - Rate_{(out)}$

with respect to x(0) = 0

$\mathbf{\dfrac{dx}{dt} = 30 - 0.015 x}$

b) Solve the initial value problem and graph both the mass of the drug and the concentration of the drug.

$\dfrac{dx}{dt} = -0.015(x - 2000)$

$\dfrac{dx}{(x - 2000)} = -0.015 \times dt$

By Using Integration Method:

$ln(x - 2000) = -0.015t + C$

$x -2000 = Ce^{(-0.015t)$

$x = 2000 + Ce^{(-0.015t)}$

However; if x(0) = 0 ;

Then

C = -2000

Therefore

$\mathbf{x = 2000 - 2000e^{-0.015t}}$

c) What is the steady-state mass of the drug in the blood?

the steady-state mass of the drug in the blood when t = infinity

$\mathbf{x = 2000 - 2000e^{-0.015 \times \infty }}$

x = 2000 - 0

x = 2000

Thus; the steady state mass of the drug is 2000 mg

d) After how many minutes does the drug mass reach 90% of its stead-state level?

After 90% of its steady state level; the mas of the drug is 90% × 2000

= 0.9 × 2000

= 1800

Hence;

$\mathbf{1800 = 2000 - 2000e^{(-0.015t)}}$

$0.1 = e^{(-0.015t)$

$ln(0.1) = -0.015t$

$t = -\dfrac{In(0.1)}{0.015}$

t = 153.5056729

t ≅ 153.51 minutes

4 0

2 years ago

What is the sum of 3/2x and 7/4x

Advocard [28]

3/2x +7/4x = ...........

3/2x(4/4)+7/4x(2/2)

12/8x+14/8x = 26/8x

26/8x = 13/4x(3.25x)

4 0

2 years ago

Read 2 more answers

Allen and Beverly thought of the same number. Allen doubled the number and added 15 to the result. He got 51. Beverly performed

pishuonlain [190]

<h2><em>Bev got 66 for her number.</em></h2>

8 0

2 years ago

Which equation correctly uses point (–2, –6) to write the equation of this line in point-slope form? Y – 6 = y minus 6 equals St

GenaCL600 [577]

Answer:

Part 1) $y+6=\frac{2}{5}(x+2)$

Part 2) $y+6=\frac{5}{2}(x+2)$

Step-by-step explanation:

The equation of the line in point slope form is

$y-y1=m(x-x1)$

we're going to analyze two cases

<em>First case</em>

$Point\ (-2,-6)$

$slope\ m=\frac{2}{5}$

substitute

$y+6=\frac{2}{5}(x+2)$

therefore

y plus 6 equals StartFraction 2 Over 5 EndFraction left-parenthesis x plus 2 right parenthesis

<em>Second case</em>

$Point\ (-2,-6)$

$slope\ m=\frac{5}{2}$

substitute

$y+6=\frac{5}{2}(x+2)$

therefore

y plus 6 equals StartFraction 5 Over 2 EndFraction left-parenthesis x plus 2 right parenthesis

5 0

2 years ago

Read 2 more answers