Algebra 2 Unit 8 lesson 2, semester b exam part 1
(47 questions)
1.D (n - 2) 180
2.D XW= YZ and XY = WZ
3.B sometimes
4. A always
5. B 45
6. C parallelogram
7. D 15 and 75
8. B 40
9. C 9
10. D 47 ft
11. C AOMN~AOJK; SAS ~
12. A AA postulate
13. A 24
14. A 6/3 miles; 6/7 miles
15. A x = 10
16. B no
17. B acute
18. C 17/2 ft
19. 27/2 + 9/2 /3
20.A 85.24
21. A sin A = 15/17 cos A = 8/17
22.A 2.0 mi
23.B 1,056 in2
24. B 25.7 ft
25. A 70 cm2
26. B 34.3 ft2
27. A 585 in2
28. B 25,000
29. D 20 in
30. B 15 m
31. D 4.2025 m2
32. B 51.8 in.2
33. A 505 m2; 519 m2
34. A 30
35. B 308 in.2
36. C 57 ft2
37. D 1,671 m2
38. B 1:4
39. C 3,463.6 in 3
40.D 795.8 ft 2
41. A 1,700 m2
42. A 8:15
43. A 52
44. D 30
45. A 43
46. A 44
47. C 28
48. B (x-3)2 + (y+2)2= 9
49. A (x+6)2 + (y-7)2 =181
I failed this for you guys so please cash app me :)
The questions are scrambled and different for everyone :(( thank you but I think everyone's gonna have to look each answer up individually
Please I need to pass this class…I don’t wanna take Algebra anymore
without the questions its a shot in the dark, im sorry! have a great summer to the ones reading this
shii i need help on this
Bro someone help out fr fr im tryna have my summer break already
someone help
Uhhhh any update I got 2 more days for this sadly
lol im doing it rn
no ones helpin us LMFAOO
uhh i still need help
Algebra sucks fr hate it
lol so were the answers wrong or yall with different connections academies
I HAVE THE ANSWERS!!
most of them, its in questions llc but its difficult to find, i didnt find it till i was partway thru the test but im here to help you so you dont end up like me,
add QUESTIONS.LLC to the begening of this link then copy and paste it into ur browser, then scroll down to find your answers
/questions/843941/2-suppose-that-the-amount-of-algae-in-a-pond-doubles-every-4-hours-if-the-pond-initially
remember, add questions.llc to the start of this link, or else it wont work.
@bucket did u ever do it?
hey Bot
Hello! How can I assist you today?
Ken has decided to rase rabbits but has been warned that the number of rabbits he will have will double every month. ken started out with 30 rabbits and the function y=30 * 2^x models the number of rabbits he will have after x months. Which graph represents this function?
The graph that represents the function y=30 * 2^x is an exponential growth curve. It would start at the point (0,30) and the y-axis would increase rapidly as x increases. You would see a steady increase in the number of rabbits over time. This graph would look similar to a curve that starts slowly, but then shoots up towards infinity as x approaches infinity.
Multiply. Give your answer in standard form.
(2n^2 + 5n +4)(2n-4)
We can use the distributive property to multiply 2n by each term in the trinomial:
```
2n(2n^2 + 5n + 4)
= 4n^3 + 10n^2 + 8n
Next, we'll multiply -4 by each term in the trinomial:
-4(2n^2 + 5n + 4)
= -8n^2 - 20n - 16
Finally, we'll combine the two products:
(2n^2 + 5n + 4)(2n-4)
= 4n^3 + 2n^2 - 12n - 16
The answer is 4n^3 + 2n^2 - 12n - 16 in standard form.
Multiply.
(3x+5)(3x-5)
We can use the formula for the square of a binomial, which is:
(a+b)(a-b) = a^2 - b^2
In this case, we can see that the binomial 3x-5 is the same as (3x) - (5), and the binomial 3x+5 is the same as (3x) + (5). So we can write:
(3x+5)(3x-5) = (3x)^2 - (5)^2
= 9x^2 - 25
Therefore, the product of (3x+5)(3x-5) is 9x^2 - 25.
the aria of a rectangular painting is given by the trinomial x^2+4x-21. what are the possible dimensions of the painting? Using factoring.
We can factor the trinomial `x^2 + 4x - 21` as follows:
```
x^2 + 4x - 21
= (x + 7)(x - 3)
```
Therefore, the dimensions of the painting are `x + 7` by `x - 3`. Since the dimensions of a painting cannot be negative, we need to find the values of `x` that make the dimensions non-negative.
Setting `x + 7 >= 0` and `x - 3 >= 0`, we get:
```
x + 7 >= 0
x >= -7
x - 3 >= 0
x >= 3
```
So the possible values of `x` are `x >= 3` and `x >= -7`. However, `x >= -7` is true for all values of `x`, so we only need to consider `x >= 3`.
Therefore, the possible dimensions of the painting are `x + 7` by `x - 3`, where `x >= 3`.
a catapult launches a boulder with an upward velocity of 148 ft/s. the height of the boulder, h, in feet after t seconds is given by the function h=-16t^2+148t+30. how long does it take the boulder to reach its maximum height? what is the boulder's maximum height? round to the nearest hundredth, if necessary.
The height of the boulder is given by the function:
h(t) = -16t^2 + 148t + 30
To find the maximum height, we need to find the vertex of the parabola that represents the height function. The x-coordinate of the vertex gives us the time at which the boulder reaches the maximum height, and the y-coordinate gives us the maximum height itself.
The x-coordinate of the vertex is given by:
t = -b / 2a
In this case, a = -16 and b = 148, so:
t = -148 / 2(-16) = 4.625
Thus, it takes the boulder 4.625 seconds to reach its maximum height.
To find the maximum height, we substitute t = 4.625 into the height function:
h(4.625) = -16(4.625)^2 + 148(4.625) + 30 = 377.625
Therefore, the boulder's maximum height is 377.63 feet (rounded to two decimal places).
a model rocket is launched from a rood into a large field. the path of the rocket can be modeled by the equation y=-0.02x^2+2.3x+6, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. how far horizontally from its starting point will the rocket land?
To find the horizontal distance traveled by the rocket, we need to find the x-intercepts of the height function. The x-intercepts occur at the points where the height of the rocket is zero, which represents the point where the rocket lands.
To find the x-intercepts, we need to solve the equation:
y = -0.02x^2 + 2.3x + 6 = 0
We can do this by factoring:
-0.02x^2 + 2.3x + 6 = 0
-0.02(x^2 - 115x - 300) = 0
(x - 15)(x - 20) = 0
Therefore, the rocket lands at a horizontal distance of either 15 meters or 20 meters from its starting point on the roof.
how many real number solutions does =-5x^(2)+8x-7 have
The equation -5x^2+8x-7 represents a quadratic function, which is a parabola.
To determine the number of real solutions, we can use the discriminant formula, which is:
d = b^2 - 4ac
where a, b, and c are the coefficients of the quadratic equation:
-5x^2 + 8x - 7 = 0
In this case, a = -5, b = 8, and c = -7. Substituting these values into the discriminant formula, we get:
d = 8^2 - 4*(-5)*(-7) = 176
Since the discriminant (176) is positive, there are two real solutions to the quadratic equation.
since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what days was the attendance the same at both plays? what was the attendance?
To find the days when attendance was the same at both plays, we need to solve the equation `y_A = y_B`, where `y_A` represents the attendance at play A and `y_B` represents the attendance at play B.
The equations for the daily attendance `y_A` and `y_B` are not provided in the question, so we cannot solve the equation. Please provide the equations for `y_A` and `y_B` so we can proceed with the solution.
since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what days was the attendance the same at both plays? what was the attendance?
play a: y=15x+76
play b: y=x^2+36x-4
To find the days when attendance was the same at both plays, we need to solve the equation `y_A = y_B`, where `y_A` represents the attendance at play A and `y_B` represents the attendance at play B:
```
y_A = y_B
15x + 76 = x^2 + 36x - 4
```
Moving all the terms to one side, we get:
```
x^2 + 21x - 80 = 0
```
We can solve this quadratic equation by factoring:
```
(x + 16)(x - 5) = 0
```
Therefore, the attendance is the same on the 16th and 5th day since opening night.
To find the attendance on those days, we can substitute the values of x into either equation:
For x = 16:
y = 15x + 76 = 15(16) + 76 = 316
For x = 5:
y = x^2 + 36x - 4 = 5^2 + 36(5) - 4 = 197
Therefore, the attendance on the 16th and 5th day since opening night was 316 and 197, respectively.
The time t required to drive a certain distance varies inversely with the speed r. if it takes 4 hours to drive the distance at 35 miles per hour, how long will it take to drive the same distance at 45 miles per hour.
If the time t required to drive a certain distance varies inversely with the speed r, we can use the formula:
t = k/r
where k is a constant of proportionality that depends on the distance.
To find the value of k, we can use the information that it takes 4 hours to drive the distance at 35 miles per hour:
4 = k/35
Solving for k, we get:
k = 140
Now we can use the value of k to find the time required to drive the same distance at 45 miles per hour:
t = 140/45
t = 3.11 hours (rounded to two decimal places)
Therefore, it will take approximately 3.11 hours to drive the same distance at 45 miles per hour.
suppose that y varies inversely with x and that y=2 when x=8. what is an equation for the inverse variation?
If y varies inversely with x, we can use the formula:
y = k/x
where k is a constant of variation.
We know that y = 2 when x = 8. Substituting these values into the formula, we get:
2 = k/8
Solving for k, we get:
k = 16
Therefore, the equation for the inverse variation is:
y = 16/x
x/x-2+x-1/x+1=-1
To solve the equation:
```
x/(x-2) + (x-1)/(x+1) = -1
```
We can begin by finding a common denominator for the two fractions on the left-hand side. The common denominator is (x-2)(x+1)(x+1), so we can rewrite the equation as:
```
x(x+1) + (x-1)(x-2) = -1(x-2)(x+1)(x+1)
```
Expanding and simplifying both sides, we get:
```
2x^2 - 2x - 3 = -x^3 - 4x^2 - 3x + 2
```
Rearranging the terms, we get a cubic equation:
```
x^3 + 6x^2 - x - 5 = 0
```
This equation is not easily solved by factoring or using the quadratic formula. One possible way to find an approximate solution is to use a numerical method, such as Newton's method or the bisection method.
Another approach is to use a graphing calculator or software to graph the left-hand side and the right-hand side of the original equation, and find the intersection points of the two graphs. This can give approximate values of the solutions.
In any case, the solution to the equation is approximately:
```
x ≈ -3.4
```
Note that this approximation is not exact and may differ slightly depending on the method used.
can you simplify that
The given equation is:
```
x/(x-2) + (x-1)/(x+1) = -1
```
First, we find a common denominator to add the two fractions on the left-hand side:
```
(x(x+1) + (x-1)(x-2)) / ((x-2)(x+1)) = -1
```
Simplifying:
```
(2x^2 - 2x - 3) / (x^2 - x - 2) = -1
```
Multiplying both sides by (x^2 - x - 2):
```
2x^2 - 2x - 3 = -x^2 + x + 2
```
Rearranging and solving for x:
```
x^3 + 6x^2 - x - 5 = 0
```
Unfortunately, this cubic equation does not have a simple factorization, so we need to use numerical or graphical methods to find the root (solution) of this equation.