The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
<h3>What is the approximation about?</h3>
From the question:
Mars: F(x) = 2/3
Therefore, If x = 15
Then:
f (15) = 2/3 ![\sqrt[8]{15}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7B15%7D)
= 16/3
= 20.7ft/s
Hence, The nearest tenth of how fast a rover will hit Mars' surface after a bounce of 15 ft in height is 20.7ft/s.
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Answer:
For 9 the answer is (4)
For 11 the answer is (5)
For 14 the answer is (4)
I cant see number 16 so maybe retake the photo thanks!
Step-by-step explanation:
9. squares have two sets of parallel lines
11. The only quadrilateral that has 3 different side lengths on this list is a trapezoid
14. It has no right angles meaning its not a square or rectangle. It has two sets of parallel sides meaning its not a trapezoid, and since it gives two different side lengths its not a rhombus since all sides muse be congruent to be a rhombus. The only one that matches is 4 a parallelogram
Answer:
28 portraits
Step-by-step explanation:
Let's first figure out how many portraits Lamy can paint in 1 week, which is his <u>unit rate</u>. To calculate this, we just have to divide the number of portraits he paints by the amount of time it takes him to paint them.
In this case, the former quantity is 84 portraits, and the latter quantity is 6 weeks, so his unit rate is 
= 14 paintings per week.
Now, we know that in 1 week, Lamy can paint 14 portraits. Therefore, since this is a <u>directly proportional relationship</u>, all we have to do to find how many portraits he can paint is 2 weeks is double the unit rate. This is because in a directly proportional relationship, if you multiply one variable by a number, you have to multiply the other by the same number to maintain equality, and here we are multiplying weeks by 2 so we need to multiply paintings by 2 as well.
Thus, Lamy can paint 14 · 2 = 28 paintings in 2 weeks.
Hope this helps!
The radius of tire is larger than radius of wheel by 5 inch
Step-by-step explanation:
We know that circumference of the circle is 2πr where “r” is the radius of the circle
Circumference refers to the dimension of the periphery of the circle. Since tires are put on the periphery of the wheel hence, we considered the circumferential aspect of the wheel.
Given-
Circumference of tires= 28π inches
2πr= 28π cancelling the common term “π” both sides
r (radius of the tires) = 14 inches
Circumference of the wheel rims= 18π
2πr= 18π cancelling the common term “π” both sides
r (radius of the tires) = 9 inches
Difference between the radius= 14-9= 5 inches
Hence, the difference between the radius of tires and the radius of the wheels is 5 inches
