Cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
<u>Solution:
</u>
Need to calculate
and then multiply the result by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
Let us first evaluate ![\sqrt[3]{1728}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1728%7D)
![\Rightarrow \sqrt[3]{1728}=\sqrt[3]{12 \times 12 \times 12}=12](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%5B3%5D%7B1728%7D%3D%5Csqrt%5B3%5D%7B12%20%5Ctimes%2012%20%5Ctimes%2012%7D%3D12)
As need to multiply 12 by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
![\Rightarrow 12 \times \sqrt[3]{14903}](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D)
On solving
, we get
![\sqrt[3]{14903}=24.608](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D%3D24.608)
![\Rightarrow 12 \times \sqrt[3]{14903}=12 \times 24.608=295.306](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D%3D12%20%5Ctimes%2024.608%3D295.306)
Hence cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
Answer:
Step-by-step explanation:
You cannot solve the problem be cause you dont know how many trees there are.
Answer:
18.87 km/hr
Step-by-step explanation:
First boat is heading North with a speed of 10 km/hr.
Second boat is heading West with a speed of 16 km/hr.
Time for which they move = 2.5 hours
To find:
The speed at which the distance is increasing between the two boats.
Solution:
Let the situation be represented by the attached diagram.
Their initial position is represented by point O from where they move towards point A and point B respectively.



We can use Pythagorean Theorem to find the distance AB.
AB is the hypotenuse of the right angled
.
According to Pythagorean theorem:

The speed at which distance is increasing between the two boats is given as:

The formula of a perimeter is:2(L+l) or
2*L+2*l
Answer:
2x-3 = 2x-5
because both sides have the same x value ( 2x) this problem has no solution
Step-by-step explanation:
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