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FTCE Mathematics 6-12 Practice Version 2 Day 2

FTCE Mathematics 6-12 Practice Version 2 Day 2

Assessment

Presentation

Mathematics

Professional Development

Easy

CCSS
HSG.GMD.A.3, HSG.C.A.2, 7.G.B.4

+17

Standards-aligned

Created by

Larry Cooper

Used 4+ times

FREE Resource

20 Slides • 54 Questions

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FTCE Mathematics 6-12 Practice Test Version 2 Day 2

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Multiple Choice

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Two lines that are parallel will have ...#76

1

negative reciprocal slopes

2

No slope

3

equal slopes

4

none of the above

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Multiple Choice

What is the equation of a line parallel to the following line that passes through the given point?

y = -4x + 6 (2, -1) #71

1

y = 1/4x + 6

2

y = -4x + 7

3

y = -4x + 2

4

y = 1/4x -1

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Multiple Choice

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y = -2/3x + 2

y = 3/2x + 9

These lines are...#72

1

Parallel

2

Perpendicular

3

Neither

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Multiple Choice

Write an equation for a line perpendicular to

y = -5x + 3 through (-5, -4). #73

1

y = -3x + 1/5

2

y = -3x - 1/5

3

y = 1/5x - 3

4

y = -1/5x - 3

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Multiple Choice

Write an equation of a line that passes through the point (-9,2) and is perpendicular to the line y = 3x - 12. #74

1

y = (-1/3)x - 1

2

y = (1/3)x - 2

3

y = -3x + 12

4

y = 3x - 1

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12

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Multiple Choice

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What is the measure of x? #81

1

12

2

15

3

6

4

9

14

Multiple Choice

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#82

1

10 ft

2

5 ft

3

8 ft

4

15 ft

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Multiple Choice

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 Find the scale factor of  ΔRPQ to ΔUST\Delta RPQ\ to\ \Delta UST .

Assume the small to large ratio. #83

1

23\frac{2}{3}  

2

22  

3

32\frac{3}{2}  

4

128\frac{12}{8}  

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Multiple Choice

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What can we conclude about MN and KL? #85

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They are parallel by Triangle Proportionality Theorem

2

They are parallel by the Converse of the Triangle Proportionality Theorem

3

They are congruent by SSS

4

They will intersect 10 units down from N

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Multiple Choice

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Which proportion could be used to prove that HJKLHJ\parallel KL  ? #86

1

KLHJ=KGLG\frac{KL}{HJ}=\frac{KG}{LG}  

2

KHKG=LGLJ\frac{KH}{KG}=\frac{LG}{LJ}  

3

HKJL=LGGK\frac{HK}{JL}=\frac{LG}{GK}  

4

GKHK=GLLJ\frac{GK}{HK}=\frac{GL}{LJ}  

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Multiple Choice

Find the interior angle on a regular polygon with 5 sides... #87

1

72o

2

108o

3

60o

4

120o

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Multiple Choice

Find the interior angle on a regular polygon with 12 sides...#88

1

36o

2

144o

3

30o

4

150o

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Today we are going to be looking at the AREA SCALE FACTOR.

Think about this question, the lengths have been multipled by 2, but what happens to the area?

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Here is another example, look at the length scale factor compared to the area scale factor.

Have you spotted the rule!?

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Multiple Choice

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The manufacturer of an ice cream company wants to change the design of its ice cream cones. The measurements of the ice cream cone are shown. If the manufacturer doubles the radius of the ice cream cone, what will be the resulting effect of the volume of the new ice cream cone? #89

1

It will increase by a factor of 2.

2

It will increase by a factor of 4.

3

It will increase by a factor of 8.

4

It will increase by a factor of 16.

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Multiple Choice

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If this circle is dilated by the scale factor of 2, what would be the new area in cm²? #90

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48 cm²

2

24 cm²

3

6 cm²

4

36 cm²

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28

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Multiple Select

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Which statements are true regarding the area of circle D? Select two options. #93

1

The area of the circle depends on the square of the radius.

2

The area of circle D is 36π cm236\pi\ cm^2 .

3

The area of circle D is 324π cm2324\pi\ cm^2 .

4

The area of the circle depends on the square of pi.

5

The area of the circle depends on the central angle.

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Multiple Choice

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The area of circle Z is.

64π ft264\pi\ ft^2 .  What is the value of r? #94

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r = 4 ft

2

r = 8 ft

3

r = 16 ft

4

r = 32 ft

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Multiple Choice

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The diagram shows one way to develop the formula for the area of a circle. Pieces of a circle with radius r are rearranged to create a shape that resembles a parallelogram.

Since the circumference of the circle can be represented by 2πr, and the area of a parallelogram is determined using A = bh, which represents the approximate area of the parallelogram-like figure? #95

1

A = (2πr)(r)

2

A = (2πr)(2r)

3

A=12(2πr)rA=\frac{1}{2}\left(2\pi r^{ }\right)r

4

A=12(2πr)r2A=\frac{1}{2}\left(2\pi r\right)r^2

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Multiple Choice

If the circumference of a circle is 10π inches, what is the area, in square inches, of the circle? #96

HINT: SOLVE FOR THE RADIUS AND USE THE RADIUS TO FIND THE AREA OF THE CIRCLE

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10π

2

25π

3

50π

4

100π

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Multiple Choice

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A circular garden with a radius of 8 feet is surrounded by a circular path with a width of 3 feet.

What is the approximate area of the path alone? Use 3.14 for π. #97

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172.70 ft2

2

178.98 ft2

3

200.96 ft2

4

379.94 ft2

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Arc Length (COPY NOTES)

Arc Length is the length of a portion of the circumference.


To find this we multiply the circumference by the fraction of the circle the arc subtends.


The fraction of the circle is determined by dividing the central angle by 360.

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Multiple Choice

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Find the arc length. Leave your answer in terms of π. S=θ2πr360° leave the π in the answer.S=\frac{\theta2\pi r}{360\degree}\ leave\ the\ \pi\ in\ the\ answer.  #99

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(7/4)π

2

(7/2)π

3

(1/8)π

4

45π

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Multiple Choice

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Find the arc length of the bold arc.  Leave π  in your answer. S=θ2πr360°S=\frac{\theta2\pi r}{360\degree}  leave π in your answer. #100

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5688π ft

2

15.8π ft

3

150.4π ft

4

54,150 ft

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Multiple Choice

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Find the arc length of the bolded arc in the picture. (Use 3.14 for π\pi  )

S=θ2πr360S=\frac{\theta\cdot2\cdot\pi\cdot r}{360}  #98

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A

2

B

3

C

4

D

40

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Multiple Choice

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What is the area of the shaded sector? #105

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69.78 ft2

2

2791.11 ft2

3

558.22 ft2

4

697.78 ft2

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Multiple Choice

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USE S=θ2πr360° to find  θfirstUSE\ S=\frac{\theta2\pi r}{360\degree}\ to\ find\ \ \theta first  Find the area of the dark blue sector shown at the left. The radius of the circle is 4 units and the length of the arc (the curved edge of the sector) measures 7.85 units. Express answer to the nearest tenth of a square unit. #106

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0.3 square units

2

15.7 square units

3

19.7 square units

4

50.3 square units

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Multiple Choice

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Find the value of x. #107

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A

2

B

3

C

4

D

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Multiple Choice

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What is the value of x? #108

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A

2

B

3

C

4

D

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Multiple Choice

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What is the value of x? #109

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A

2

B

3

C

4

D

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Multiple Choice

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What is the measure of angle A? #115

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34°

2

180°

3

112°

4

79°

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Multiple Choice

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Solve for x and y. #116

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x=85⁰ y = 150⁰

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x = 85⁰ y = 50⁰

3

x = 95⁰ y = 150⁰

4

x = 95⁰ y = 50⁰

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Multiple Choice

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Find m∠R. #117

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24⁰

2

41⁰

3

67⁰

4

139⁰

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Multiple Choice

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Jenny used a compass and a straight edge to make this construction. Which statement best describes her construction? #119

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She is constructing a parallel line to PQ.

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She is copying line segment PQ.

3

She is constructing the perpendicular bisector to PQ.

4

She is constructing an angle bisector.

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Multiple Choice

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Name the construction. #120

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Corresponding angles

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Parallel line to a point not on the line

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Perpendicular line to a point not on the line

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Angle bisector

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Multiple Choice

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Find the volume of the figure. Round to the nearest tenth. #123

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80 cm³

2

40 cm²

3

80 cm2

4

40 cm³

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Multiple Choice

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What is the volume(L*W*H) of this rectangular prism with a width of 2 and a length of 7 and a height of 4? #125

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56 cm3

2

28 cm3

3

14 cm3

4

8 cm3

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Multiple Choice

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Find the volume of the cylinder. V=πr2h where r is the radius and h is the height.V=\pi r^2h\ where\ r\ is\ the\ radius\ and\ h\ is\ the\ height.  #129

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4239

2

1696

3

1130

4

565

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Multiple Choice

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Find the volume of the figure. V=(length)(width)(height)3V=\frac{\left(length\right)\left(width\right)\left(height\right)}{3}  #130

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126 cm3

2

907.5 cm3

3

605 cm3

4

55 cm3

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Multiple Choice

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Find the volume of the composite figure.

Vtotal=Vpyramid+VprismV_{total}=V_{pyramid}+V_{prism}  Length and width of the pyramid is 12 ft. #131

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2400 ft3

2

1920 ft3

3

4128 ft3

4

2112 ft3

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Multiple Choice

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A right square pyramid has a height of 15 cm. and a slant height of 25 cm.

What is the volume of the pyramid? #132

1

2400 square cm.

2

6000 square cm.

3

8000 square cm.

4

24000 square cm.

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Multiple Choice

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Find the volume of the Sphere. Use 3.14 for pi. V=4πr33V=\frac{4\pi r^3}{3}  #136

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337.3 ft3

2

137.2 ft3

3

274.4 ft³

4

205.8 ft³

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Multiple Choice

Change the following equation to standard form

x2+y2-4x+12y-8=0. #138

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(x-2)2+(y-6)2=48

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(x-2)2+(y+6)2=48

3

(x-6)2+(y-2)2=48

4

(x+2)2+(y-6)2=48

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Multiple Choice

What is the equation of the circle, with center (-2,-7) and radius 3 units? #139

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(x + 2)2 + (y + 7)2 = 3

2

(x - 2)2 + (y - 7)2 = 9

3

(x + 2)2 + (y + 7)2 = 9

4

(x - 2)2 + (y - 7)2 = 3

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Multiple Choice

The endpoints of a diameter of a circle are (3,1) and (9,5).Find the equation of the circle? Hint: In this problem, we are not given the center or radius however we can find the length of the diameter using the distance formula and then divide it by 2. The center is simply the midpoint of the given points. d=(x2x1)2+(y2y1)2d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}   Use Midpoint Formula to find the center  (x1+x22,y1+y22)Use\ Midpoint\ Formula\ to\ find\ the\ center\ \ \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)   #140

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(x−6)2+(y−3)2=13

2

(x−3)2+(y−6)2=132

3

(x−5)2+(y−9)2=13

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(x−9)2+(y−5)2=169/4

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Multiple Choice

Convert 150⁰ to radians. #144

MULTIPLY BY (π/180)

1

5π/6

2

3π/4

3

7π/6

4

2π/3

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Match

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Match the following angles measured in radians to degrees. #146

90°90\degree  

   180°180\degree

30°30\degree

45°45\degree

60°60\degree

π2\frac{\pi}{2}  

π\pi

π6\frac{\pi}{6}

π4\frac{\pi}{4}

π3\frac{\pi}{3}

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Multiple Choice

Point P(7, -24) is a point on the terminal side of the angle θ\theta . Find the exact value of sine θ\theta .

#149

1

2425-\frac{24}{25}  

2

725\frac{7}{25}  

3

247-\frac{24}{7}  

4

2425\frac{24}{25}  

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Multiple Choice

Point R(3, 4) is a point on the terminal side of the angle θ\theta . Find the exact value of tangent θ\theta . #150

1

34\frac{3}{4}  

2

43-\frac{4}{3}  

3

45\frac{4}{5}  

4

43\frac{4}{3}  

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Multiple Choice

Which of the following formulas shows the Law of Cosines? #151

1

c2 = a2 + b2 - 4ac + cosA

2

c2 = a2 - b2 - 2abcosC

3

c2 = a2 + b2 - 2abcosC

4

(cos A)/a = (cos B)/b

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Multiple Choice

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What is the correct set up to find angle H?

 #152

1

  cos H = 1121321524(11)(15)\cos\ H\ =\ \frac{11^2-13^2-15^2}{4\left(11\right)\left(15\right)}  

2

cos H = 112+132+1522(13)(15)\cos\ H\ =\ \frac{11^2+13^2+15^2}{2\left(13\right)\left(15\right)}   

3

  cos H = 1121321522(13)(15)\cos\ H\ =\ \frac{11^2-13^2-15^2}{-2\left(13\right)\left(15\right)}  

4

H2 = 112+132+1522(13)(15)cos11H^2\ =\ 11^2+13^2+15^2-2\left(13\right)\left(15\right)\cos11   

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Multiple Choice

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#153

1

60

2

66

3

76

4

96

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Multiple Choice

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Identify the Law of Sines. #155

1

sin2x+cos2x=1\sin^2x+\cos^2x=1

2

sin(A)a=sin(B)b=sin(C)c\frac{\sin\left(A\right)}{a}=\frac{\sin\left(B\right)}{b}=\frac{\sin\left(C\right)}{c}

3

sin(A)sin(B)sin(C)=1\sin\left(A\right)\sin\left(B\right)\sin\left(C\right)=1

4

Soh Cah Toa

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Multiple Choice

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Use the Law of Sines to solve for BC. #156

1

33

2

24

3

12

4

29

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Multiple Select

Solve for 0≤x≤2π

2sin2x - sinx = 0 #158

1

0, π, 2π

2

π/2, 3π/2

3

π/6, 5π/6

4

π/3, 2π/3

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Multiple Choice

Solve for 0≤x≤2π

3sin2x + sinx - 4 = 0 #159

1

0, π

2

π/2

3

π/6, 5π/6

4

3π/2

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Multiple Choice

Solve for 0≤x≤2π

tanx = 1 #161

1

π/2, π

2

π/4, 3π/4

3

π/4, 5π/4

4

3π/4, 7π/4

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FTCE Mathematics 6-12 Practice Test Version 2 Day 2

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