1. assume that a surveyor stands at the top of a mountain that is "h" feet tall. if the distance (in feet) that he can see is defined by d = 3200.2 sqrt(h), then answer the following. (a) how far can the surveyor see from the top of a 2000-foot mountain? (b) how tall is the mountain, if the surveyor can see 15 miles? (note: 1 mile equals 5280 feet.)
2. suppose the altitude of a rising hot-air balloon is given by h = 0.04 t2 + 2t, where "t" is the time in seconds after the balloon leaves the ground. how long will it take for the balloon to reach an altitude of 200 feet?
3. devon tosses a horseshoe at a stake 30 feet away. the horseshoe lands no more than 3 feet from the stake. write an absolute value inequality that represents the range of distances that the horseshoe travels.
how in gods name do you do these problems?

1a) 143,112.9 feet.

1b) 612.06 feet.

2) 50 seconds

3) |d -30| ≤ 3

Step-by-step explanation:

The first two problems give you a function defined in terms of a variable and ask you to find some data that has to do either with the variable or the function.

Remember that a function f(x) = y is a relation between sets X and Y that assigns an element of Y to each element of X.

When we define a function between two variables, we say that there is a relation between them and we can calculate one another in terms of the other variable.

1. Assume that a surveyor stands at the top of a mountain that is "h" feet tall. If the distance (in feet) that he can see is defined by d = 3200.2 SQRT(h).

Here you have a function where the distance is defined in terms of the height: d =3200.2 √h (both of them in feet)

In a function we can "plug in" a value of the variable and apply it to the function and it will give us the result of the other variable.

So, we have:

(a) How far can the surveyor see from the top of a 2000-foot mountain?

Here the problem is giving us the height h = 2000 feet and he asks us how far he can see (distance) so we're going to substitute h in our expression and solve for d.

d =3200.2 √h

d = 3200.2 √2000  

d= 3200.2 (44.72)

d = 143,112.9

Therefore he can see 143,112.9 feet.

b) How tall is the mountain, if the surveyor can see 15 miles? (Note: 1 mile equals 5280 feet.)

Here the problem asks us for the height h, but first we are going to convert the 15 miles into feet: If 1 mile equals 5280 feet, then 15 miles equals

5280(15) = 79200 feet.

Then d = 79200 feet

So we have:

d =3200.2 √h

79200 = 3200.2 √h

79200/3200.2 = √h

24.74 = √h

24.74² = h

612.06 = h

Thus, the mountain is 612.06 feet tall

2. Suppose the altitude of a rising hot-air balloon is given by h = 0.04 t2 + 2t, where "t" is the time in seconds after the balloon leaves the ground. How long will it take for the balloon to reach an altitude of 200 feet?

This problem is similar to the one before, we have height expressed in terms of t (time in seconds). The problem asks us how long will it take for the balloon to reach 200 feet. So h = 200

h = 0.04t² + 2t

200 = 0.04t²+ 2t

since this is a quadratic expression we are going to equal to zero

0.04t² + 2t -200 = 0 now we are going to factorize the 0.04 (in other words, divide every term by 0.04)

0.04 (t² + 50t - 5000) = 0 by solving by factoring the trinomial we get:

0.04 (t +100) (t - 50) = 0

Therefore t + 100 = 0 or t -50 = 0

t + 100 = 0 ⇒ t = - 100 but time cannot be in negative (it cannot take -100 seconds for the balloon to reach some altitude)

t - 50 = 0 ⇒ t = 50.

Therefore, this is our answer and it will take the balloon 50 seconds to reach a height of 200 feet.

3. Devon tosses a horseshoe at a stake 30 feet away. The horseshoe lands no more than 3 feet from the stake. Write an absolute value inequality that represents the range of distances that the horseshoe travels.

So Devon tosses a horseshoe at a stake 30 feet away, the horseshoe will land no more than 3 feet away from the stake, so it can travel from 27 feet (and it lands 3 feet before the stake) to 33 feet (and it lands 3 feet after the stake).

If we call d the distance the horseshoe travels we would get

|d - 30| ≤ 3 *Remember that the absolute value of a number always gives you the positive value, for example |3| = 3, |-3| = 3 *

so with these inequality we can only use numbers d such that when we subtract 30 from them, we will get 3.

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value

225x²+4=0

225x²=0-4

225x²=-4

x²=-4/225

x=±√-4/225