MA 1025 MOD 1 QUESTION 6-20
Shaungio74
MA1025MOD1HWNEW.docxQuestion1: Score 1/3
|
Classify the number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 00 on the number line?
−6.210735-6.210735 is |
|
Your response |
Correct response |
|
positive |
negative |
|
Auto graded |
Grade: 0/1.0 |
|
and |
|
Your response |
Correct response |
|
rational |
rational |
|
Auto graded |
Grade: 1/1.0 |
|
. It lies to the |
|
Your response |
Correct response |
|
right |
left |
|
Auto graded |
Grade: 0/1.0 |
|
of 0.0. |
Total grade: 0.0×1/3 + 1.0×1/3 + 0.0×1/3 = 0% + 33% + 0%
Feedback:
e. −6.210735-6.210735 is a terminating decimal so it is negative and rational. It lies to the left of 00.
Question2: Score 1/3
|
Classify the number as either positive or negative and as either rational or irrational. Does the number lie to the left or the right of 00 on the number line? 5–√252 is |
|
Your response |
Correct response |
|
negative |
positive |
|
Auto graded |
Grade: 0/1.0 |
|
and |
|
Your response |
Correct response |
|
irrational |
irrational |
|
Auto graded |
Grade: 1/1.0 |
|
. It lies to the |
|
Your response |
Correct response |
|
left |
right |
|
Auto graded |
Grade: 0/1.0 |
|
of 0.0. |
Total grade: 0.0×1/3 + 1.0×1/3 + 0.0×1/3 = 0% + 33% + 0%
Feedback:
d. 5–√252 is positive and irrational. It lies to the right of 00.
Question3: Score 1/1
|
According to the U.S. Mint, the diameter of a dollar is 1.0431.043 inches. The circumference of the dollar would be the diameter multiplied by ππ.
Is the circumference of a dollar a whole number, a rational number, or an irrational number?
|
|
Your response |
Correct response |
|
Irrational Number Feedback: Correct. |
|
|
Auto graded |
Grade: 1/1.0 |
|
Hint |
Penalty |
|
|
Hint |
0.0 |
|
Total grade: 1.0×1/1 = 100%
Feedback:
The circumference of a dollar is 1.043π1.043π. Since ππ is neither a whole number, nor a rational number, we conclude the circumference is an irrational number.
Question4: Score 1/1
|
Identify the number as rational, irrational, whole, or natural. Choose the answer that is the most descriptive.
−14−14
|
|
Your response |
Correct response |
|
Rational number Feedback: Correct. |
|
|
Auto graded |
Grade: 1/1.0 |
|
|
Total grade: 1.0×1/1 = 100%
Feedback:
The number −14−14 is a rational number.
Question5: Score 0/1
|
Identify the number as rational, irrational, whole, or natural. Choose the answer that is the most descriptive.
5–√5
|
|
Your response |
Correct response |
|
Natural number Feedback: Incorrect. |
Irrational number |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
The number 5–√5 is an irrational number.
Question6: Score 0/1
|
In chemistry the volume for a certain gas is given by V=25TV=25T, where VV is measured in cc and TT is temperature in °C°C. If the temperature varies between 90°C90°C and 110°C110°C , find the set of volume values.
Enter the exact answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
Do not enter any commas in your answer.
The range of the volume is |
|
Your response |
Correct response |
|
{x | 90≤x≤110} {x | 90≤x≤110} |
[2250, 2750] |
|
Auto graded |
Grade: 0/1.0 |
|
cc. |
Total grade: 0.0×1/1 = 0%
Feedback:
We are given the volume for a certain gas, V=25TV=25T, and TT varies between 90°C90°C and 110°C110°C. We need to find the range of the volume.
Express the range of temperature as an inequality, and find the range for the volume.
90≤T≤11090≤T≤110
(25)90≤(25)T≤(25)1102590≤25T≤25110
Multiply by 2525.
2,250≤V≤2,7502,250≤V≤2,750
[2,250,2,750]2,250,2,750
Convert to interval notation.
Therefore, the range of the volume is [2,250,2,750]2,250,2,750.
Question7: Score 0/1
|
Describe all the xx-values within or including a distance of 44 units from the number 88.
Enter your answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
44 88 |
[4, 12] |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
We want the distance between xx and 88 to be less than or equal to 44. We can draw a number line, such as in the figure below, to represent the condition to be satisfied.
44 44
88
The distance from xx to 88 can be represented using an absolute value symbol, |x−8|x−8. Write the values of xx that satisfy the condition as an absolute value inequality.
|x−8|≤4x−8≤4
We need to write two inequalities as there are always two solutions to an absolute value equation.
x−8x−8
≤4≤4
and
x−8x−8
≥−4≥−4
xx
≤12≤12
xx
≥4≥4
If the solution set is x≤12x≤12 and x≥4x≥4, then the solution set is an interval including all real numbers between and including 44 and 1212.
So |x−8|≤4x−8≤4 is equivalent to [4,12]4,12 in interval notation.
Question8: Score 0/1
|
Write the set in interval notation. {x|−4<x≤8}x|−4<x≤8
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
-4 8 |
(-4, 8] |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
Use a parentheses on the left of −4−4 and bracket after 88: (−4,8]−4,8.
Question9: Score 0/1
|
Solve the inequality involving absolute value.
|x+4|≥−9x+4≥−9
Enter the exact answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
{0 |
(-infinity, infinity) |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
|x+4|x+4
≥−9≥−9
x+4x+4
≤9≤9
x+4x+4
≥−9≥−9
xx
≤5≤5
xx
≥−13≥−13
If the solution set is x≤5x≤5 and x≥−13x≥−13, then the solution set is an interval including all real numbers. So the solution set is (−∞,∞)−∞,∞.
Question10: Score 0/1
|
Solve the inequality.
4x−10≤64x−10≤6
Enter the exact answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
-0.17 0.25 |
(-infinity, 4] |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
4x−104x−10
≤6≤6
4x4x
≤16≤16
(14)4x144x
≤16(14)≤1614
Multiply by 1414.
xx
≤4≤4
Express the inequality in interval notation.
(−∞,4]−∞,4
Question11: Score 0/1
|
Solve the inequality.
−4(x−1)+2>2x−3−3x−4x−1+2>2x−3−3x
Enter the exact answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
(-8, 0.25) |
(-infinity, 3) |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
−4(x−1)+2−4x−1+2
>2x−3−3x>2x−3−3x
−4x+4+2−4x+4+2
>−x−3>−x−3
−4x+6−4x+6
>−x−3>−x−3
Collect like terms.
−3x−3x
>−9>−9
Isolate the variable term.
xx
<3<3
Dividing both sides by −3−3 reverses the inequality.
Express the inequality in interval notation.
(−∞,3)−∞,3
Question12: Score 0/1
|
Solve the inequality involving absolute value.
|x−3|+4≥10x−3+4≥10
Enter the exact answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
[0, 0.2]∪ [0.3, 0.9] |
(-infinity, -3] U [9, infinity) |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
|x−3|+4x−3+4
≥10≥10
|x−3|x−3
≥6≥6
x−3x−3
≤−6≤−6
x−3x−3
≥6≥6
xx
≤−3≤−3
xx
≥9≥9
Express the inequality in interval notation.
(−∞,−3]∪[9,∞)−∞,−3∪9,∞
Question13: Score 0/1
|
Solve the compound inequality.
2x−8<−142x−8<−14 or 7x+2≥97x+2≥9
Enter the exact answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
[−6,-1)∪ {0, 0}∪ (1,0.07} |
(-infinity, -3) U [1, infinity) |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
Solve both inequalities:
2x−82x−8
<−14<−14
or
7x+27x+2
≥9≥9
2x2x
<−6<−6
7x7x
≥7≥7
xx
<−3<−3
xx
≥1≥1
Therefore, the solution set is x<−3x<−3 and x≥1x≥1 or in interval notation (−∞,−3)∪[1,∞)−∞,−3∪1,∞.
Question14: Score 0/1
|
Solve the equation for xx.
5x+5=3x−85x+5=3x−8
Enter the exact answer. Improper fractions are acceptable solutions.
x=x=
|
|
Your response |
Correct response |
|
8/17 |
-13/2 |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
Apply standard algebraic operations.
5x+55x+5
=3x−8=3x−8
5x+5−55x+5−5
=3x−8−5=3x−8−5
Subtract 55 from both sides.
5x5x
=3x−13=3x−13
Combine like terms.
2x2x
=−13=−13
Place xx- terms on one side and simplify.
xx
=−132=−132
Divide both sides by 22.
Question15: Score 0/1
|
Solve the equation for xx.
x+27−x−16=2x+27−x−16=2 Enter the exact answer.
x=x= |
|
Your response |
Correct response |
|
2 |
-65 |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
We have two denominators: 77 and 66. The LCD is 4242.
Next, multiply the whole equation (both sides of the equal sign) by 4242.
42(x+27−x−16)42x+27−x−16
=(2)42=242
7⋅6(x+27)−7⋅6(x−16)7⋅6x+27−7⋅6x−16
=84=84
Distribute 4242.
6(x+2)−7(x−1)6x+2−7x−1
=84=84
Denominators cancel out.
6x+12−7x+76x+12−7x+7
=84=84
Use the distributive property.
−x+19−x+19
=84=84
Combine like terms.
−x−x
=65=65
Subtract 1919 from both sides.
xx
=−65=−65
Divide both sides by −1−1.
Question16: Score 0/1
|
Solve the equation for x.
4(x+3)−42=184x+3−42=18
x=x= |
|
Your response |
Correct response |
|
[0.0.5, 0.3] |
12 |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
4(x+3)−424x+3−42
=18=18
4(x+3)−42+424x+3−42+42
=18+42=18+42
Add 4242 to both sides.
4(x+3)4x+3
=60=60
4(x+3)44x+34
=604=604
Divide both sides by 44.
x+3x+3
=15=15
x+3−3x+3−3
=15−3=15−3
Subtract 33 from both sides.
xx
=12=12
Question17: Score 0/1
|
Solve for xx.
6(x+3)−27=x+66x+3−27=x+6
Enter the exact answer.
x=x= |
|
Your response |
Correct response |
|
0.03 |
3 |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
Apply standard algebraic properties.
6(x+3)−276x+3−27
=x+6=x+6
6x+18−276x+18−27
=x+6=x+6
Apply the distributive property.
6x−96x−9
=x+6=x+6
Combine like terms.
5x5x
=15=15
Place xx- terms on one side and simplify.
xx
=155=155
Multiply both sides by 1515, the reciprocal of 55.
xx
=3=3
Question18: Score 0/2
|
Solve for xx. 13+3x=141513+3x=1415
Enter the exact answer.
x=x= |
|
Your response |
Correct response |
|
46699.3 |
5 |
|
Auto graded |
Grade: 0/1.0 |
|
State all xx-values that are excluded from the solution set.
The field below accepts a list of numbers or formulas separated by semicolons (e.g. 2;4;62;4;6 or x+1;x−1x+1;x−1). The order of the list does not matter.
x≠x≠ |
|
Your response |
Correct response |
|
0; 1415; 41513;141500 |
0 |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/2 + 0.0×1/2 = 0% + 0%
Feedback:
Only one value is excluded from the solution set, 00. We have three denominators: 33, xx, and 1515. No factoring is required. The LCD is (3)(5)x35x.
Next, multiply the whole equation (both sides of the equal sign) by (3)(5)x35x.
(3)(5)x(13+3x)35x13+3x
=(1415)(3)(5)x=141535x
(3)(5)x(13)+(3)(5)x(3x)35x13+35x3x
=(1415)(3)(5)x=141535x
(5)x(1)+(3)(5)(3)5x1+353
=(14)x=14x
Denominators cancel out.
5x+455x+45
=14x=14x
9x9x
=45=45
Collect like terms.
xx
=459=459
Multiply by 1919, the reciprocal of 99.
xx
=5=5
Question19: Score 0/1
|
Solve the inequality.
−3x+4>x−20−3x+4>x−20
Enter the exact answer in interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
(0, 0) U (∞) |
(-infinity, 6) |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
Solving this inequality is similar to solving an equation up until the last step.
−3x+4−3x+4
>x−20>x−20
−4x+4−4x+4
>−20>−20
Move variable terms to one side of the inequality.
−4x−4x
>−24>−24
Isolate the variable term.
xx
<6<6
Dividing both sides by −4−4 reverses the inequality.
The solution set is given by the interval (−∞,6)−∞,6.
Question20: Score 0/1
|
Solve the compound inequality.
−5<2x+3≤30−5<2x+3≤30
Enter the exact answer in interval notation. Improper fractions are acceptable in the interval notation.
To enter ∞∞, type infinity. To enter ∪∪, type U.
|
|
Your response |
Correct response |
|
(-4, 13.5) U [14, 35] |
(-4,27/2] |
|
Auto graded |
Grade: 0/1.0 |
|
|
Total grade: 0.0×1/1 = 0%
Feedback:
Leave the compound inequality intact, and perform solving procedures on the three parts at the same time.
−5−5
<2x+3≤30<2x+3≤30
−8−8
<2x≤27<2x≤27
Isolate the variable term, and subtract 33 from all three parts.
−4−4
<x≤272<x≤272
Divide through all three parts by 22.
The solution set is given by the interval (−4,272]−4,272.
0
