25 Questions - Algebra 2, College level.

(See attached file for full problem description with equations)

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1. The diameter of the Milky Way disc is approximately 9  1020 meters. How long does it take light, traveling at 1016 m/year to travel across the diameter of the Milky Way?
Time = distance/speed = 9  1020 m/1016 m/year = 9*104 years

2. Divide. =

3. Multiply. Write the answer in scientific notation.
(2.4  10-5)(4  10-4) = 9.6*10-9

4. Simplify. =

5. Multiply
(a2 + 2ab - b2)(a2 - 7ab + b2) = a2(a2 - 7ab + b2)+2ab(a2 - 7ab + b2)-b2(a2 - 7ab + b2)
= a4 - 7a3b + a2b2 + 2a3b - 14a2b2 + 2ab3- a2b2 + 7ab3 - b4
= a4 - 5a3b - 14a2b2 + 9ab3 - b4

6. Factor completely. 12x3 - 3xy2 = 3x(4x2 - y2) = 3x(2x + y)(2x - y)

7. Factor. 8m4n - 16mn4 = 8mn(m3 - 2n3)

8. The product of two consecutive positive even integers is 288. Find the integers.
a(a+2) = 288
or a2 + 2a - 288 = 0
Therefore, a =
Taking positive value, a = 17 - 1 = 16
The other integer, a + 2 = 18

9. Factor completely. 3x2 - 5xy + 3x - 5y = x(3x - 5y) + 1(3x - 5y) = (x + 1)(3x - 5y)

10. Factor completely. 5x2 + 33x + 18 = 5x2 + 3x + 30x + 18 = x(5x + 3) + 6(5x + 3)
= (5x + 3)(x + 6)

11. Write in simplest form. =

12. Solve.
Or
Or
Or 6x - 15 = 7x + 14
Therefore, x = -29

13. Multiply. =

14. During rush hour, Fernando can drive 40 miles using the side roads in the same time that it takes to travel 30 miles on the freeway. If Fernando's rate on the side roads is 7 mi/h faster than his rate on the freeway, find his rate on the side roads.
Let Fernando's speed on the freeway be x mi/h.

Or 1 + 7/x = 1+ 1/3
Therefore, x = 21 mi/h
Therefore, x + 7 = 28 mi/h

15. Solve. =
Therefore, 8(x-5) = (14-x)(x+5)
Or 8x - 40 = -x2 + 70 + 9x
Or x2 - x - 110 = 0
Or (x - 11)(x + 10) = 0 implies that x = -10, 11

16. Simplify. =

17. Find the length x. Express your answer in simplified radical form.
x =
X =

18. Find the altitude (or height) of the triangle.

Height h =

19. Evaluate , if possible.
=

20. Use a calculator to approximate to the nearest hundredth.
= -7.87

21. Match the graph with its equation.

The

The graph passes through the origin. Therefore, (0,0) should satisfy the equation of the graph, that is, the equation should not contain any constant term. This rules out B and D.
Also, the point (1,0) lies on the graph. Plugging in this point in C gives: y = -12 + 2*1 = 1 which is not zero. This rules out C. Therefore, A is the correct answer.

A) y = -x2 + x B) y = -x2 + 1 C) y = -x2 + 2x D) y = -x2 - 1

22. Solve by completing the square.
x2 - 10x + 24 = 0
Or x2 - 10x = -24
Adding 25 to both sides, x2 - 10x + 25 = 25-24
Or x2 - 2*5x + 52 = 1
Or (x -5)2 = 12
Therefore, x - 5 = 1 or -1
Therefore, x = 5 + 1 or 5 - 1 = 6, 4

23. A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?
Let the width of the walkway be x ft
Then, (30 + 2x)(40 + 2x) = 1800
Or 2(15 + x)*2(20 + x) = 1800
Or (15 + x)(20 + x) = 450
Or x2 + 35x + 300 = 450
Therefore, x2 + 35x - 150 = 0
Therefore, x =
(Taking just the positive sign)

24. Solve by using the quadratic formula. 3x2 = 11x + 4
Or 3x2 - 11x - 4 = 0
Therefore, x =

25. Identify the axis of symmetry, create a suitable table of values, then sketch the graph (including the axis of symmetry).

y = -x2 + 3x - 3
or y + 3 = -x2 + 3x
Adding -9/4 to both sides, y + 3 - 9/4 = -x2 + 3x - 9/4
Or (y + ¾) = -(x - 3/2)2
Therefore, the vertex is (3/2, -3/4)
Axis of symmetry is: x = -3/2

x y= -x2+3x-3
-8 37
-7.5 30.75
-7 25
-6.5 19.75
-6 15
-5.5 10.75
-5 7
-4.5 3.75
-4 1
-3.5 -1.25
-3 -3
-2.5 -4.25
-2 -5
-1.5 -5.25
-1 -5
-0.5 -4.25
0 -3
0.5 -1.25
1 1
1.5 3.75
2 7
2.5 10.75
3 15
3.5 19.75
4 25
4.5 30.75
5 37
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