Initially started 20 feet from the ground .
Went uphill 83 feet.
So: 20 + 83 = 103. The coaster is 103 ft above at this point.

Went down 42 feet
So: 103 - 42 = 61.

Went up 128 feet
So: 61 + 128 = 189

Finally, went down 90 feet
So: 189 - 90 = 99

99 ft above ground.

3 0

3 years ago

Read 2 more answers

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago

What is the slope of a line parallel to the line whose equation is 5x+y=-9

Rzqust [24]

Answer:

slope = - 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c (m is the slope and c the y- intercept )

Given

5x + y = - 9 ( subtract 5x from both sides )

y = - 5x - 9 ← in slope- intercept form

with slope m = - 5

Parallel lines have equal slopes, thus

The slope of a parallel line to the given equation is m = - 5

8 0

3 years ago