All HSPT Math Resources
Example Questions
Example Question #2085 : Hspt Mathematics
Begin at the point on the coordinate plane and move four units to the right. Next, move six units up. Where are you now?
Moving four units to the right increases the -coordinate by 4; moving six units up increases the -coordinate by 6. The current location is
, or .
Example Question #1492 : Concepts
On the coordinate plane, how do you get from to ?
Move left ten units and up two units
Move down ten units and left two units
Move up ten units and right two units
Move right ten units and down two units
Move right ten units and down two units
The -coordinate changes by units; therefore, you must move ten units in a positive horizontal direction - right.
The -coordinate changes by units; therefore, you must move two units in a negative vertical direction - down.
The correct response is to move right ten units and down two units.
Example Question #1491 : Concepts
How far from the origin is the point on the coordinate plane?
Apply the distance formula, substituting the -coordinates for and and the -coordinates for and :
Example Question #2092 : Hspt Mathematics
Two lines on the coordinate plane are perpendicular and intersect at the origin. One has slope . Give the equation of the other.
A line perpendicular to a line with slope has as its slope the opposite of the reciprocal of ; this number is . The line passes through the origin, which is the point ; this is also the -intercept. In slope-intercept form, its equation can be found by substituting in the equation:
This can be rewritten in standard form as follows:
Example Question #2091 : Hspt Mathematics
A line with slope passes through the origin and the point . Evaluate .
The origin is . In the following slope formula, set
,
and solve for :
Example Question #1494 : Concepts
The lines of which two of the following equations are perpendicular to each other?
(I)
(II)
(III)
II and III
I and II
No two of the equations given are represented by lines that are perpendicular to each other.
I and III
I and II
and are represented by a horizontal line and a vertical line, respectively; therefore, their graphs are perpendicular.
As for the third equation, , this is represented by a line with slope , as seen below:
The equation is in slope-intercept form, and the slope of the line is the coefficient of , which is . It is perpendicular to a line with slope ; however, the horizontal line has slope 0, and the vertical line has undefined slope.
The correct choice is therefore I and II.
Example Question #2092 : Hspt Mathematics
Start at the origin on the coordinate plane. Move up units, right units, down units, and right units, and you will be at the point .
Evaluate .
Since you move right units in the second step and right units in the fourth, you will move a total of units right; this means the -coordinate will be , which is equal to 72.
Since you move up units in the first step and down units in the third, you will move a total of units up; this means the -coordinate will be , which is equal to 46.
We can form a system of equations and solve by adding, as follows:
Example Question #2091 : Hspt Mathematics
What is the -intercept of the graph of the function
The graph has no -intercept.
The -intercept of the graph of a function is the point at which it intersects the -axis - that is, at which . This point is , so evaluate :
The -intercept is .
Example Question #1 : Cylinders
A right circular cylinder has a height of 41 in. and a lateral area (excluding top and bottom) 512.5π in2. What is the area of its bases?
None of the other answers
312.5 in2
78.125π in2
156.25 in2
39.0625π in2
78.125π in2
The lateral area (not including its bases) is equal to the circumference of the base times the height of the cylinder. Think of it like a label that is wrapped around a soup can. Therefore, we can write this area as:
A = h * π * d or A = h * π * 2r = 2πrh
Now, substituting in our values, we get:
512.5π = 2 * 41*rπ; 512.5π = 82rπ
Solve for r by dividing both sides by 82π:
6.25 = r
From here, we can calculate the area of a base:
A = 6.252π = 39.0625π
NOTE: The question asks for the area of the bases. Therefore, the answer is 2 * 39.0625π or 78.125π in2.
Example Question #3 : How To Find The Surface Area Of A Cylinder
The number of square units in the surface area of a right circular cylinder is equal to the number of cubic units in its volume. If r and h represent the length in units of the cylinder's radius and height, respectively, which of the following is equivalent to r in terms of h?
r = 2h/(h – 2)
r = h/(2h – 2)
r = 2h2 + 2
r = h2/(h + 2)
r = h2 + 2h
r = 2h/(h – 2)
We need to find expressions for the surface area and the volume of a cylinder. The surface area of the cylinder consists of the sum of the surface areas of the two bases plus the lateral surface area.
surface area of cylinder = surface area of bases + lateral surface area
The bases of the cylinder will be two circles with radius r. Thus, the area of each will be πr2, and their combined surface area will be 2πr2.
The lateral surface area of the cylinder is equal to the circumference of the circular base multiplied by the height. The circumferece of a circle is 2πr, and the height is h, so the lateral area is 2πrh.
surface area of cylinder = 2πr2 + 2πrh
Next, we need to find an expression for the volume. The volume of a cylinder is equal to the product of the height and the area of one of the bases. The area of the base is πr2, and the height is h, so the volume of the cylinder is πr2h.
volume = πr2h
Then, we must set the volume and surface area expressions equal to one another and solve for r in terms of h.
2πr2 + 2πrh = πr2h
First, let's factor out 2πr from the left side.
2πr(r + h) = πr2h
We can divide both sides by π.
2r(r + h) = r2h
We can also divide both sides by r, because the radius cannot equal zero.
2(r + h) = rh
Let's now distribute the 2 on the left side.
2r + 2h = rh
Subtract 2r from both sides to get all the r's on one side.
2h = rh – 2r
rh – 2r = 2h
Factor out an r from the left side.
r(h – 2) = 2h
Divide both sides by h – 2
r = 2h/(h – 2)
The answer is r = 2h/(h – 2).