Answer:
The total cost of redecoration is £66, that is less than £75. Tenant can decorate the room for the budget set.
Step-by-step explanation:
Lets start calculating the areas of shorter and longer walls.
The area of shorter wall is
2.65m*2.98m=7.897m^2.
The area of longer wall is
3.94m*2,65m=10.441m^2.
Together the area of one long and one short wall is
7.897m^2+10.441m^2=18.338m^2.
The area to be painted ( one short and one long wall) is 18.338m^2 that is less than 30m^2, and hence only one tin of paint is needed (costs £16).
If there is assumed that the sheets of wallpaper must be whole from floor to ceiling, the calculations are following.
We get from one roll
10m/2.65m=3.77 sheets; we can use only 3 of them
The width of the walls is 3,94m+2,98m=6,92m,
and division gives 6.92m/0.52m=13,3... .
Hence it requires 14 sheets, that is 5 rolls of wallpaper.
Wallpaper costs 5*£10=£50 and altogether it is needed £50+£16=£66 to redecorate the room.
9514 1404 393
Answer:
- base: 2.18 m
- height: 7.35 m
Step-by-step explanation:
Let b represent the base of the triangle in meters. Then the height is ...
h = 3 +2b
and the area is ...
A = 1/2bh
8 = 1/2(b)(3 +2b)
16 = 3b +2b^2 . . . . . . . multiply by 2 and eliminate parentheses
2(b^2 +3/2b + (9/16)) = 16 + 9/8 . . . . . . complete the square
2(b +3/4)^2 = 17.125
(b +3/4)^2 = 8.5625 . . . . . divide by 2
b + 0.75 = √8.5625 ≈ 2.9262
b = -0.75 +2.9262 = 2.1762
h = 3 +2b = 7.3523
The base is about 2.18 meters and the height is about 7.35 meters.
Answer:
Mean = 1.9
Standard deviation = 0.6
Step-by-step explanation:
The Mean is calculate by the formula:
⇒ 
Thus Mean to the nearest tenth is 1.9
Standard Deviation is the square root of sum of square of the distance of observation from the mean.
where, 
is mean of the distribution.
Putting all values in the formula, We get
Standard Deviation = 0.597 ≈ 0.6
Answer:
The answer is below
Step-by-step explanation:
The linear model represents the height, f(x), of a water balloon thrown off the roof of a building over time, x, measured in seconds: A linear model with ordered pairs at 0, 60 and 2, 75 and 4, 75 and 6, 40 and 8, 20 and 10, 0 and 12, 0 and 14, 0. The x axis is labeled Time in seconds, and the y axis is labeled Height in feet. Part A: During what interval(s) of the domain is the water balloon's height increasing? (2 points) Part B: During what interval(s) of the domain is the water balloon's height staying the same? (2 points) Part C: During what interval(s) of the domain is the water balloon's height decreasing the fastest? Use complete sentences to support your answer. (3 points) Part D: Use the constraints of the real-world situation to predict the height of the water balloon at 16 seconds.
Answer:
Part A:
Between 0 and 2 seconds, the height of the balloon increases from 60 feet to 75 feet at a rate of 7.5 ft/s
Part B:
Between 2 and 4 seconds, the height stays constant at 75 feet.
Part C:
Between 4 and 6 seconds, the height of the balloon decreases from 75 feet to 40 feet at a rate of -17.5 ft/s
Between 6 and 8 seconds, the height of the balloon decreases from 40 feet to 20 feet at a rate of -10 ft/s
Between 8 and 10 seconds, the height of the balloon decreases from 20 feet to 0 feet at a rate of -10 ft/s
Hence it fastest decreasing rate is -17.5 ft/s which is between 4 to 6 seconds.
Part D:
From 10 seconds, the balloon is at the ground (0 feet), it continues to remain at 0 feet even at 16 seconds.
A. 4x + 9 = 34
<u> - 9 - 9</u>
<u>4x</u> = <u>25</u>
4 4
x = 6.25
B. (x - 4)(x + 2) = 0
x - 4 = 0 U x + 2 = 0
<u> + 4 + 4</u> <u> - 2 - 2</u>
x = 4 x = -2
C. 2x² - 6x + 4 = 0
2(x²) - 2(3x) + 2(2) = 0
<u>2(x² - 3x + 2)</u> = <u>0</u>
2 2
x² - 3x + 2 = 0
x = <u>-(-3) +/- √((-3)² - 4(1)(2))</u>
2(1)
x = <u>3 +/- √(9 - 8)</u>
2
x = <u>3 +/- √(1)
</u> 2<u>
</u> x =<u> 3 +/- 1
</u> 2
x = <u>3 + 1</u> U x = <u>3 - 1</u>
2 2
x = <u>4</u> x = <u>2</u>
2 2
x = 2 x = 1
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