Honors Analysis
6.5: Right Triangle Applications
Q1.) A lunar lander blasts off from the surface of the moon. The function modeling its distance from the surface is , where y represents the number of meters above the surface the lander is after x seconds. Including correct units, what is the average rate of change of the lander from x = 1 to x = 5 seconds?
Q2.) Use the trapezoidal rule with three trapezoids to approximate , that is, the area bounded by the graph, the x-axis, and the lines x = 1 and x = 4.
Q3.) What is the minimum value of the function ? Show all work.
Q4.) Use the quadratic formula to solve (and be careful with signs!!!): .
Q5.) Factor completely and give all solutions:
Give the formula for the surface area of a:
Q6.)ConeQ7.)CylinderQ8.)Sphere
Give the formula for the volume of a:
Q9.)CylinderQ10.)ConeQ11.)Sphere
1. Given that the base of the pyramid shown is 10 x 102.)
and PR = 12, calculate:
A)the length of the slant height (PS):
B)the surface area of the pyramid
C)the volume of the pyramid
3.4. Find each area:
5.)
6.)7.) What is the side length of an equilateral
triangle with an area of sq. cm?
8.) 9.) What is the area of a regular octagon with
10 cm sides? You may round to the nearest tenth.
10. A cone has a diameter 18 in and a height11.)
of 12 in.
A) What is the slant height?
B) Determine the lateral area and total area.
C) Calculate the volume.
12.)
13.) A prism with a regular pentagonal base with
6 cm sides has an altitude of 10 cm. Calculate the
volume to the nearest tenth.
14.A pyramid has a regular hexagonal base 15.)
with 6 ft sides and an altitude of 8 ft.
A) Determine the volume of the pyramid.
B) Calculate the lateral surface area of the
pyramid.
C) What is the total surface area?
16.)
17.)
Honors Analysis
Enrichment Lesson #3: Solving Equations With Rational Exponents
To solve the equation , first divide out the coefficient, resulting in . At this point, you would like to solve for x. The goal is to force the exponent of x to be one, which can be done by taking both sides of the function to the reciprocal power since exponents are multiplied when a power is raised to a power, and multiplying by the reciprocal results in a product of 1. In this case, take both sides to the2/3 power: , resulting in .
*Note: Make sure addition/subtraction outside parentheses is handled BEFORE using exponents (otherwise, you may have to use FOIL and things get messy!)
Solve for x:
E1.)E2.)E3.)E4.)
In general, you can’t solve equations in which the exponent is a variable using basic algebra. For example, if you wanted to solve , most people assume dividing by 5 will work. But if you think about exponent rules, when you divide by 5 you are really dividing by , which results in:
Of course, this equation isn’t any easier to solve than the original. So clearly this approach doesn’t work. In most cases, you will need logarithms to solve equations with exponents that are variables. In special cases, however, you can rewrite equations using the same number as the base. For example:
Both 25 and 125 can be written as powers of x: 52 and 53, which can be substituted in place for the 25 and 125. Then exponent rules can be applied (multiply exponents when taking a power to a power):
At this point, it is known that . Since the bases are the same, the exponents must be the same, so set them equal: 12 = 3x +15, so x = -1.
Solve each equation for x by writing both sides of the equation as powers of the same number:
E5.)E6.)E7.)