2. Another wooden plank is leaning against an outside wall of a building. The bottom of the plank is 9 ft from the wall. The length of the plank is 41 ft. Find each of the following values, and show all your work.(a) Find the value of x. Round to the nearest tenth of a degree.
(b) Find the height above the ground where the plank touches the wall. Round to the nearest foot.
Please someone help
Answers
Given that the distance of the ground is 9ft.
Length of the plank is 41 ft
In order to get the angle formed between the ground and the foot of the plank,
cos (theta) = ground / plank
cos (theta) = 9 / 41
cos (theta) = 0.21951...
theta = cos^-1(0.21951...)
theta = 77.3 degrees
(b) Find the height above the ground where the plank touches the wall. Round to the nearest tenth of a foot.
Height = square root (plank^2 - ground^2)
Height = square root (41^2 - 8^2)
Height = square root (1600)
Height = 40
(A) 77.1 Degrees
(B) 40 feet
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Answers
Answers
Final answer:
The equation representing the elevation of Grand Lake after x number of days, given a decrease of 3 inches per day, is Elevation = 648 - 0.25x, where 0.25 is the daily loss rate in feet and x represents the number of days.
Explanation:
The subject of this question is the application of linear equations to real-world problems, specifically in the context of environmental changes. Given the initial elevation of Grand Lake and the daily rate of water level decrease, we can write an equation that represents the elevation of the lake's surface after x days.
The initial height of the lake is 648 feet. Every day, the lake elevation drops by 3 inches. However, our rate is given in inches and the initial elevation is in feet. We should convert the rate from inches to feet. Since there are 12 inches in a foot, 3 inches is equal to 3/12 = 0.25 feet.
So after one day, the lake would be at 648 - 0.25 = 647.75 feet. After two days, the lake would then be at 648 - 2(0.25) = 647.50 feet, and so on. In general, after x days, the elevation of the lake would be 648 - x(0.25)
Therefore, the equation we are looking for is: Elevation = 648 - 0.25x
Learn more about Linear Equations here:
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To Prove: x = 2
1.
Division property of equality
A. 12 - x = 20 - 5 x
2.
Subtraction property of equality
B. 12 + 4 x = 20
3. Addition property of equality
C. 4 x = 8
4. given
D. x=2
Answers
(4) Given: (A) 12 - x = 20 - 5x
Then, add 5x to both sides of the equation,
(3) Addition property of equality: (B) 12 + 4x = 20
Then, subtract 12 from both sides of the equation
(2) Subtraction property of equality: (C) 4x = 8
Lastly, divide both sides by 4
(1) Division property of equality : (D) x = 2
Answers
Answers
Answer:
see the picture for the answer
Answer:
Step-by-step explanation:
F(x) = 2x - 1; G(x) = 3x + 2;
F[G(x)] - F(x).= F(3x+2) -F(x) = 2(3x+2)-1 -(2x-1) = 6x +4-1-2x+1
F[G(x)] - F(x).= 4x+4
Answers
Answer:
A box-and-whisker plot shows the scores on a math exam for two classes.
Class A Class B
1) Minimum value 65 56
2) Lower quartile 66 79
3) Median 81 91
4) Upper quartile 89 94
5) Maximum value 95 100
First quartile also known as Lower quartile is the number below which lies the 25 percent of the bottom data.
Second quartile or Median divides the range in the middle and has 50 percent of the data below it.
Third quartile also known as upper quartile has 75 percent of the data below it and the top 25 percent of the data above it.
Interquartile Range = Upper quartile - Lower quartile,
For class A
Interquartile range = 89 -66 = 23
For class B
Interquartile range = 94 -79 = 15
The interquartile ranges tell you about the two classes that Class B has more consistent scores
Class A Class B
1) minimum value 65 56
2) 1st quartile 66 79
3) median 81 91
4) 3rd quartile 89 94
5) maximum value 95 100
Interquartile range = 3rd Quartile - 1st Quartile
Class A: 89 - 66 = 23
Class B: 94 - 79 = 15
IQR indicates the spread of the of the scores over the median. Class A has a spread that his more than 50% of the scores of Class B.