cpm precalculus homework answers

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• Core Connections Integrated I, 2013
• Core Connections Algebra 1, 2013
• Core Connections Geometry, 2013
• Core Connections Algebra 2, 2013
• Core Connections Integrated I, 2014
• Core Connections Integrated II, 2015
• Core Connections: Course 1
• Core Connections: Course 2
• Core Connections: Course 3
• Core Connections Integrated III, 2015

8.1 Non-right Triangles: Law of Sines

α = 98 ∘ a = 34.6 β = 39 ∘ b = 22 γ = 43 ∘ c = 23.8 α = 98 ∘ a = 34.6 β = 39 ∘ b = 22 γ = 43 ∘ c = 23.8

β ≈ 5.7° , γ ≈ 94.3° , c ≈ 101.3 β ≈ 5.7° , γ ≈ 94.3° , c ≈ 101.3

about 8.2 8.2 square feet

8.2 Non-right Triangles: Law of Cosines

a ≈ 14.9 , a ≈ 14.9 , β ≈ 23.8° , β ≈ 23.8° , γ ≈ 126.2° . γ ≈ 126.2° .

α ≈ 27.7° , α ≈ 27.7° , β ≈ 40.5° , β ≈ 40.5° , γ ≈ 111.8° γ ≈ 111.8°

Area = 552 square feet

about 8.15 square feet

8.3 Polar Coordinates

( x , y ) = ( 1 2 , − 3 2 ) ( x , y ) = ( 1 2 , − 3 2 )

r = 3 r = 3

x 2 + y 2 = 2 y x 2 + y 2 = 2 y or, in the standard form for a circle, x 2 + ( y − 1 ) 2 = 1 x 2 + ( y − 1 ) 2 = 1

8.4 Polar Coordinates: Graphs

The equation fails the symmetry test with respect to the line θ = π 2 θ = π 2 and with respect to the pole. It passes the polar axis symmetry test.

Tests will reveal symmetry about the polar axis. The zero is ( 0 , π 2 ) , ( 0 , π 2 ) , and the maximum value is ( 3 , 0 ) . ( 3 , 0 ) .

The graph is a rose curve, n n even

Rose curve, n n odd

8.5 Polar Form of Complex Numbers

| z | = 50 = 5 2 | z | = 50 = 5 2

z = 3 ( cos ( π 2 ) + i sin ( π 2 ) ) z = 3 ( cos ( π 2 ) + i sin ( π 2 ) )

z = 2 ( cos ( π 6 ) + i sin ( π 6 ) ) z = 2 ( cos ( π 6 ) + i sin ( π 6 ) )

z = 2 3 − 2 i z = 2 3 − 2 i

z 1 z 2 = − 4 3 ; z 1 z 2 = − 3 2 + 3 2 i z 1 z 2 = − 4 3 ; z 1 z 2 = − 3 2 + 3 2 i

z 0 = 2 ( cos ( 30° ) + i sin ( 30° ) ) z 0 = 2 ( cos ( 30° ) + i sin ( 30° ) )

z 1 = 2 ( cos ( 120° ) + i sin ( 120° ) ) z 1 = 2 ( cos ( 120° ) + i sin ( 120° ) )

z 2 = 2 ( cos ( 210° ) + i sin ( 210° ) ) z 2 = 2 ( cos ( 210° ) + i sin ( 210° ) )

z 3 = 2 ( cos ( 300° ) + i sin ( 300° ) ) z 3 = 2 ( cos ( 300° ) + i sin ( 300° ) )

8.6 Parametric Equations

x ( t ) = t 3 − 2 t y ( t ) = t x ( t ) = t 3 − 2 t y ( t ) = t

y = 5 − 1 2 x − 3 y = 5 − 1 2 x − 3

y = ln x y = ln x

x 2 4 + y 2 9 = 1 x 2 4 + y 2 9 = 1

y = x 2 y = x 2

8.7 Parametric Equations: Graphs

The graph of the parametric equations is in red and the graph of the rectangular equation is drawn in blue dots on top of the parametric equations.

8.8 Vectors

3 u = 〈 15 , 12 〉 3 u = 〈 15 , 12 〉

u = 8 i − 11 j u = 8 i − 11 j

v = 34 cos ( 59° ) i + 34 sin ( 59° ) j v = 34 cos ( 59° ) i + 34 sin ( 59° ) j

Magnitude = 34 34

θ = tan − 1 ( 5 3 ) = 59.04° θ = tan − 1 ( 5 3 ) = 59.04°

8.1 Section Exercises

The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.

When the known values are the side opposite the missing angle and another side and its opposite angle.

A triangle with two given sides and a non-included angle.

β = 72° , a ≈ 12.0 , b ≈ 19.9 β = 72° , a ≈ 12.0 , b ≈ 19.9

γ = 20° , b ≈ 4.5 , c ≈ 1.6 γ = 20° , b ≈ 4.5 , c ≈ 1.6

b ≈ 3.78 b ≈ 3.78

c ≈ 13.70 c ≈ 13.70

one triangle, α ≈ 50.3° , β ≈ 16.7° , a ≈ 26.7 α ≈ 50.3° , β ≈ 16.7° , a ≈ 26.7

two triangles, γ ≈ 54.3° , β ≈ 90.7° , b ≈ 20.9 γ ≈ 54.3° , β ≈ 90.7° , b ≈ 20.9 or γ ′ ≈ 125.7° , β ′ ≈ 19.3° , b ′ ≈ 6.9 γ ′ ≈ 125.7° , β ′ ≈ 19.3° , b ′ ≈ 6.9

two triangles, β ≈ 75.7° , γ ≈ 61.3° , b ≈ 9.9 β ≈ 75.7° , γ ≈ 61.3° , b ≈ 9.9 or β ′ ≈ 18.3° , γ ′ ≈ 118.7° , b ′ ≈ 3.2 β ′ ≈ 18.3° , γ ′ ≈ 118.7° , b ′ ≈ 3.2

two triangles, α ≈ 143.2° , β ≈ 26.8° , a ≈ 17.3 α ≈ 143.2° , β ≈ 26.8° , a ≈ 17.3 or α ′ ≈ 16.8° , β ′ ≈ 153.2° , a ′ ≈ 8.3 α ′ ≈ 16.8° , β ′ ≈ 153.2° , a ′ ≈ 8.3

no triangle possible

A ≈ 47.8° A ≈ 47.8° or A ′ ≈ 132.2° A ′ ≈ 132.2°

370.9 370.9

29.7° 29.7°

x = 76.9° or x = 103.1° x = 76.9° or x = 103.1°

110.6° 110.6°

A ≈ 39.4 , C ≈ 47.6 , B C ≈ 20.7 A ≈ 39.4 , C ≈ 47.6 , B C ≈ 20.7

430.2 430.2

A D ≈ 13.8 A D ≈ 13.8

A B ≈ 2.8 A B ≈ 2.8

L ≈ 49.7 , N ≈ 56.3 , L N ≈ 5.8 L ≈ 49.7 , N ≈ 56.3 , L N ≈ 5.8

The distance from the satellite to station A A is approximately 1716 miles. The satellite is approximately 1706 miles above the ground.

2.6 ft

5.6 km

371 ft

5936 ft

24.1 ft

19,056 ft 2

445,624 square miles

8.65 ft 2

8.2 Section Exercises

two sides and the angle opposite the missing side.

s s is the semi-perimeter, which is half the perimeter of the triangle.

The Law of Cosines must be used for any oblique (non-right) triangle.

not possible

B ≈ 45.9° , C ≈ 99.1° , a ≈ 6.4 B ≈ 45.9° , C ≈ 99.1° , a ≈ 6.4

A ≈ 20.6° , B ≈ 38.4° , c ≈ 51.1 A ≈ 20.6° , B ≈ 38.4° , c ≈ 51.1

A ≈ 37.8° , B ≈ 43.8 , C ≈ 98.4° A ≈ 37.8° , B ≈ 43.8 , C ≈ 98.4°

177.56 in 2

292.4 miles

8.3 Section Exercises

For polar coordinates, the point in the plane depends on the angle from the positive x- axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations.

Determine θ θ for the point, then move r r units from the pole to plot the point. If r r is negative, move r r units from the pole in the opposite direction but along the same angle. The point is a distance of r r away from the origin at an angle of θ θ from the polar axis.

The point ( − 3 , π 2 ) ( − 3 , π 2 ) has a positive angle but a negative radius and is plotted by moving to an angle of π 2 π 2 and then moving 3 units in the negative direction. This places the point 3 units down the negative y -axis. The point ( 3 , − π 2 ) ( 3 , − π 2 ) has a negative angle and a positive radius and is plotted by first moving to an angle of − π 2 − π 2 and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y -axis.

( − 5 , 0 ) ( − 5 , 0 )

( − 3 3 2 , − 3 2 ) ( − 3 3 2 , − 3 2 )

( 2 5 , 0.464 ) ( 2 5 , 0.464 )

( 34 , 5.253 ) ( 34 , 5.253 )

( 8 2 , π 4 ) ( 8 2 , π 4 )

r = 4 csc θ r = 4 csc θ

r = s i n θ 2 c o s 4 θ 3 r = s i n θ 2 c o s 4 θ 3

r = 3 cos θ r = 3 cos θ

r = 3 sin θ cos ( 2 θ ) r = 3 sin θ cos ( 2 θ )

r = 9 sin θ cos 2 θ r = 9 sin θ cos 2 θ

r = 1 9 cos θ sin θ r = 1 9 cos θ sin θ

x 2 + y 2 = 4 x x 2 + y 2 = 4 x or ( x − 2 ) 2 4 + y 2 4 = 1 ; ( x − 2 ) 2 4 + y 2 4 = 1 ; circle

3 y + x = 6 ; 3 y + x = 6 ; line

y = 3 ; y = 3 ; line

x y = 4 ; x y = 4 ; hyperbola

x 2 + y 2 = 4 ; x 2 + y 2 = 4 ; circle

x − 5 y = 3 ; x − 5 y = 3 ; line

( 3 , 3 π 4 ) ( 3 , 3 π 4 )

( 5 , π ) ( 5 , π )

r = 6 5 cos θ − sin θ r = 6 5 cos θ − sin θ

r = 2 sin θ r = 2 sin θ

r = 2 cos θ r = 2 cos θ

x 2 + y 2 = 16 x 2 + y 2 = 16

y = x y = x

x 2 + ( y + 5 ) 2 = 25 x 2 + ( y + 5 ) 2 = 25

( 1.618 , − 1.176 ) ( 1.618 , − 1.176 )

( 10.630 , 131.186° ) ( 10.630 , 131.186° )

( 2 , 3.14 ) o r ( 2 , π ) ( 2 , 3.14 ) o r ( 2 , π )

A vertical line with a a units left of the y -axis. 

A horizontal line with a a units below the x -axis.

8.4 Section Exercises

Symmetry with respect to the polar axis is similar to symmetry about the x x -axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line θ = π 2 θ = π 2 is similar to symmetry about the y y -axis.

Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at θ = 0 , π 2 , θ = 0 , π 2 , π π and  3 π 2 , 3 π 2 , and sketch the graph.

The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.

symmetric with respect to the polar axis

symmetric with respect to the polar axis, symmetric with respect to the line θ = π 2 , θ = π 2 , symmetric with respect to the pole

no symmetry

symmetric with respect to the pole

one-loop/dimpled limaçon

inner loop/two-loop limaçon

Archimedes’ spiral

They are both spirals, but not quite the same.

Both graphs are curves with 2 loops. The equation with a coefficient of θ θ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to 4 π 4 π to get a better picture.

When the width of the domain is increased, more petals of the flower are visible.

The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole.

The graphs are spirals. The smaller the coefficient, the tighter the spiral.

( 4 , π 3 ) , ( 4 , 5 π 3 ) ( 4 , π 3 ) , ( 4 , 5 π 3 )

( 3 2 , π 3 ) , ( 3 2 , 5 π 3 ) ( 3 2 , π 3 ) , ( 3 2 , 5 π 3 )

( 0 , π 2 ) , ( 0 , π ) , ( 0 , 3 π 2 ) , ( 0 , 2 π ) ( 0 , π 2 ) , ( 0 , π ) , ( 0 , 3 π 2 ) , ( 0 , 2 π )

( 8 4 2 , π 4 ) , ( 8 4 2 , 5 π 4 ) ( 8 4 2 , π 4 ) , ( 8 4 2 , 5 π 4 ) and at θ = 3 π 4 , θ = 3 π 4 , 7 π 4 7 π 4 since r r is squared

8.5 Section Exercises

a is the real part, b is the imaginary part, and i = − 1 i = − 1

Polar form converts the real and imaginary part of the complex number in polar form using x = r cos θ x = r cos θ and y = r sin θ . y = r sin θ .

z n = r n ( cos ( n θ ) + i sin ( n θ ) ) z n = r n ( cos ( n θ ) + i sin ( n θ ) ) It is used to simplify polar form when a number has been raised to a power.

14.45 14.45

4 5 cis ( 333.4° ) 4 5 cis ( 333.4° )

2 cis ( π 6 ) 2 cis ( π 6 )

7 3 2 + i 7 2 7 3 2 + i 7 2

− 2 3 − 2 i − 2 3 − 2 i

− 1.5 − i 3 3 2 − 1.5 − i 3 3 2

4 3 cis ( 198° ) 4 3 cis ( 198° )

3 4 cis ( 180° ) 3 4 cis ( 180° )

5 3 cis ( 17 π 24 ) 5 3 cis ( 17 π 24 )

7 cis ( 70° ) 7 cis ( 70° )

5 cis ( 80° ) 5 cis ( 80° )

5 cis ( π 3 ) 5 cis ( π 3 )

125 cis ( 135° ) 125 cis ( 135° )

9 cis ( 240° ) 9 cis ( 240° )

cis ( 3 π 4 ) cis ( 3 π 4 )

3 cis ( 80° ) , 3 cis ( 200° ) , 3 cis ( 320° ) 3 cis ( 80° ) , 3 cis ( 200° ) , 3 cis ( 320° )

2 4 3 cis ( 2 π 9 ) , 2 4 3 cis ( 8 π 9 ) , 2 4 3 cis ( 14 π 9 ) 2 4 3 cis ( 2 π 9 ) , 2 4 3 cis ( 8 π 9 ) , 2 4 3 cis ( 14 π 9 )

2 2 cis ( 7 π 8 ) , 2 2 cis ( 15 π 8 ) 2 2 cis ( 7 π 8 ) , 2 2 cis ( 15 π 8 )

3.61 e − 0.59 i 3.61 e − 0.59 i

− 2 + 3.46 i − 2 + 3.46 i

− 4.33 − 2.50 i − 4.33 − 2.50 i

8.6 Section Exercises

A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, x = f ( t ) x = f ( t ) and y = f ( t ) . y = f ( t ) .

Choose one equation to solve for t , t , substitute into the other equation and simplify.

Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.

y = − 2 + 2 x y = − 2 + 2 x

y = 3 x − 1 2 y = 3 x − 1 2

x = 2 e 1 − y 5 x = 2 e 1 − y 5 or y = 1 − 5 l n ( x 2 ) y = 1 − 5 l n ( x 2 )

x = 4 log ( y − 3 2 ) x = 4 log ( y − 3 2 )

x = ( y 2 ) 3 − y 2 x = ( y 2 ) 3 − y 2

y = x 3 y = x 3

( x 4 ) 2 + ( y 5 ) 2 = 1 ( x 4 ) 2 + ( y 5 ) 2 = 1

y 2 = 1 − 1 2 x y 2 = 1 − 1 2 x

y = x 2 + 2 x + 1 y = x 2 + 2 x + 1

y = ( x + 1 2 ) 3 − 2 y = ( x + 1 2 ) 3 − 2

y = − 3 x + 14 y = − 3 x + 14

y = x + 3 y = x + 3

{ x ( t ) = t y ( t ) = 2 sin t + 1 { x ( t ) = t y ( t ) = 2 sin t + 1

{ x ( t ) = t + 2 t y ( t ) = t { x ( t ) = t + 2 t y ( t ) = t

{ x ( t ) = 4 cos t y ( t ) = 6 sin t ; { x ( t ) = 4 cos t y ( t ) = 6 sin t ; Ellipse

{ x ( t ) = 10 cos t y ( t ) = 10 sin t ; { x ( t ) = 10 cos t y ( t ) = 10 sin t ; Circle

{ x ( t ) = − 1 + 4 t y ( t ) = − 2 t { x ( t ) = − 1 + 4 t y ( t ) = − 2 t

{ x ( t ) = 4 + 2 t y ( t ) = 1 − 3 t { x ( t ) = 4 + 2 t y ( t ) = 1 − 3 t

yes, at t = 2 t = 2

answers may vary: { x ( t ) = t − 1 y ( t ) = t 2  and  { x ( t ) = t + 1 y ( t ) = ( t + 2 ) 2 { x ( t ) = t − 1 y ( t ) = t 2  and  { x ( t ) = t + 1 y ( t ) = ( t + 2 ) 2

answers may vary: , { x ( t ) = t y ( t ) = t 2 − 4 t + 4  and  { x ( t ) = t + 2 y ( t ) = t 2 { x ( t ) = t y ( t ) = t 2 − 4 t + 4  and  { x ( t ) = t + 2 y ( t ) = t 2

8.7 Section Exercises

plotting points with the orientation arrow and a graphing calculator

The arrows show the orientation, the direction of motion according to increasing values of t . t .

The parametric equations show the different vertical and horizontal motions over time.

There will be 100 back-and-forth motions.

Take the opposite of the x ( t ) x ( t ) equation.

The parabola opens up.

{ x ( t ) = 5 cos t y ( t ) = 5 sin t { x ( t ) = 5 cos t y ( t ) = 5 sin t

a = 4 , a = 4 , b = 3 , b = 3 , c = 6 , c = 6 , d = 1 d = 1

a = 4 , a = 4 , b = 2 , b = 2 , c = 3 , c = 3 , d = 3 d = 3

The y y -intercept changes.

y ( x ) = − 16 ( x 15 ) 2 + 20 ( x 15 ) y ( x ) = − 16 ( x 15 ) 2 + 20 ( x 15 )

{ x ( t ) = 64 t cos ( 52 ° ) y ( t ) = − 16 t 2 + 64 t sin ( 52 ° ) { x ( t ) = 64 t cos ( 52 ° ) y ( t ) = − 16 t 2 + 64 t sin ( 52 ° )

approximately 3.2 seconds

1.6 seconds

8.8 Section Exercises

lowercase, bold letter, usually u , v , w u , v , w

They are unit vectors. They are used to represent the horizontal and vertical components of a vector. They each have a magnitude of 1.

The first number always represents the coefficient of the i , i , and the second represents the j . j .

〈 7 , − 5 〉 〈 7 , − 5 〉

7 i − 3 j 7 i − 3 j

− 6 i − 2 j − 6 i − 2 j

u + v = 〈 − 5 , 5 〉 , u − v = 〈 − 1 , 3 〉 , 2 u − 3 v = 〈 0 , 5 〉 u + v = 〈 − 5 , 5 〉 , u − v = 〈 − 1 , 3 〉 , 2 u − 3 v = 〈 0 , 5 〉

− 10 i – 4 j − 10 i – 4 j

− 2 29 29 i + 5 29 29 j − 2 29 29 i + 5 29 29 j

− 2 229 229 i + 15 229 229 j − 2 229 229 i + 15 229 229 j

− 7 2 10 i + 2 10 j − 7 2 10 i + 2 10 j

| v | = 7.810 , θ = 39.806 ° | v | = 7.810 , θ = 39.806 °

| v | = 7.211 , θ = 236.310° | v | = 7.211 , θ = 236.310°

〈 4 , 1 〉 〈 4 , 1 〉

v = − 7 i + 3 j v = − 7 i + 3 j

3 2 i + 3 2 j 3 2 i + 3 2 j

i − 3 j i − 3 j

x = 7.13 x = 7.13 pounds, y = 3.63 y = 3.63 pounds

x = 2.87 x = 2.87 pounds, y = 4.10 y = 4.10 pounds

4.635 miles, 17.764° N of E

17 miles. 10.318 miles

Distance: 2.868. Direction: 86.474° North of West, or 3.526° West of North

4.924°. 659 km/hr

( 0.081 , 8.602 ) ( 0.081 , 8.602 )

21.801°, relative to the car’s forward direction

parallel: 16.28, perpendicular: 47.28 pounds

19.35 pounds, 231.54° from the horizontal

5.1583 pounds, 75.8° from the horizontal

Review Exercises

Not possible

C = 120° , a = 23.1 , c = 34.1 C = 120° , a = 23.1 , c = 34.1

distance of the plane from point A : A : 2.2 km, elevation of the plane: 1.6 km

b = 71.0° , C = 55.0° , a = 12.8 b = 71.0° , C = 55.0° , a = 12.8

( 0 , 2 ) ( 0 , 2 )

( 9.8489 , 203.96° ) ( 9.8489 , 203.96° )

r = 8 r = 8

x 2 + y 2 = 7 x x 2 + y 2 = 7 x

y = − x y = − x

symmetric with respect to the line θ = π 2 θ = π 2

cis ( − π 3 ) cis ( − π 3 )

2.3 + 1.9 i 2.3 + 1.9 i

60 cis ( π 2 ) 60 cis ( π 2 )

3 cis ( 4 π 3 ) 3 cis ( 4 π 3 )

25 cis ( 3 π 2 ) 25 cis ( 3 π 2 )

5 cis ( 3 π 4 ) , 5 cis ( 7 π 4 ) 5 cis ( 3 π 4 ) , 5 cis ( 7 π 4 )

x 2 + 1 2 y = 1 x 2 + 1 2 y = 1

{ x ( t ) = − 2 + 6 t y ( t ) = 3 + 4 t { x ( t ) = − 2 + 6 t y ( t ) = 3 + 4 t

y = − 2 x 5 y = − 2 x 5

• { x ( t ) = ( 80 cos ( 40° ) ) t y ( t ) = − 16 t 2 + ( 80 sin ( 40° ) ) t + 4 { x ( t ) = ( 80 cos ( 40° ) ) t y ( t ) = − 16 t 2 + ( 80 sin ( 40° ) ) t + 4
• The ball is 14 feet high and 184 feet from where it was launched.
• 3.3 seconds

− 3 10 10 − 3 10 10 i − 10 10 − 10 10 j

Magnitude: 3 2 , 3 2 , Direction: 225° 225°

Practice Test

α = 67.1° , γ = 44.9° , a = 20.9 α = 67.1° , γ = 44.9° , a = 20.9

1712 miles 1712 miles

( 1 , 3 ) ( 1 , 3 )

y = − 3 y = − 3

− 5 2 + i 5 3 2 − 5 2 + i 5 3 2

4 cis ( 21° ) 4 cis ( 21° )

2 2 cis ( 18° ) , 2 2 cis ( 198° ) 2 2 cis ( 18° ) , 2 2 cis ( 198° )

y = 2 ( x − 1 ) 2 y = 2 ( x − 1 ) 2

−4 i − 15 j

2 13 13 i + 3 13 13 j 2 13 13 i + 3 13 13 j

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Access for free at https://openstax.org/books/precalculus/pages/1-introduction-to-functions

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• Book title: Precalculus
• Publication date: Oct 23, 2014
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Home > PC3 > Chapter 1 > Lesson 1.1.1 > Problem 1-5

All of the graphs used in your Team Sort were functions. Describe what it means for a graph to be a “function.” Sketch two non-functions and explain why these two examples are not functions.      

A function must have exactly one output for every input.

Think about what happens to functions rotated 90º or some of the conic functions.

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CPM Student Tutorials

Other resources.

• Student: CPM eBooks (Student Version)
• Student: eWorkspace
• Parent: eBook Support
• Trouble Shooting
• Creating Desmos eTools
• Creating CPM eTools
• Algebra Videos
• Problem Solving Videos
• Statistics Videos
• TI-84 Graphing Calculator
• Student: Presentation Tools
• CC Course 1 eTools
• CC Course 2 eTools
• CC Course 3 eTools
• CC Algebra eTools
• CC Geometry eTools
• CC Algebra 2 eTools
• CC Integrated I eTools
• CC Integrated II eTools
• CC Integrated III eTools
• Pre-Calc. w/Trig
• Making Connections 1
• Making Connections 2
• Algebra Connections
• Geometry Connections
• Algebra 2 Connections
• Foundations For Algebra 1
• Foundations For Algebra 2
• Calculus 1.2.1: 1-15 & 1-18 Student eTools
• Calculus 1.2.2: 1-31 & 1-32 Student eTools