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3 years ago

Find the cube root of 2700

Mathematics

2 answers:

cupoosta [38]3 years ago

5 0

13.92 or we can break it down to 3 and cube root of 100

Liono4ka [1.6K]3 years ago

3 0

It’s about 13.925. I rounded up

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Find the common ratio for 4,2, and 1

svetoff [14.1K]

Answer:

4:2:1 is common ratio for 4,2, and 1

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3 years ago

Which of the following is the algebraic form for the verbal statement shown: "12 more than the product 4 and a number, n" - A. n

katrin2010 [14]

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then the answer is B

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3 years ago

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Van wants to build a square field with a fence around the outside. If he has 104 feet of

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Answer:

Step-by-step explanation:

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3 years ago

Factorise polynomial<br> 27y3 + 125z3<br><br>

tresset_1 [31]

Answer:

$27y^{3} +125z^{3} =(3y)^{3} +(5z)^{3}$

Apply the formula for sum of cubes: $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$

Therefore:

$\left(3y\right)^3+\left(5z\right)^3=\left(3y+5z\right)\left(3^2y^2-3\cdot \:5yz+5^2z^2\right)$

$=\left(3y+5z\right)\left(9y^2-15yz+25z^2\right)$

6 0

3 years ago

A hiker is hiking in a valley. The height of the valley is h(x,y)=4x2+y2 where x and y are the east-west and north-south distanc

Ainat [17]

Answer:

A. $\frac{\partial{h}}{\partial{t}}=0$

Step-by-step explanation:

A. The problems asked for 2 ways to solve it, expanding the equation with the substitution x(t)=2 cos(t) and y(t)=4 sin(t) to differentiate it . The other way is by chain rule.

Expanding and differentiating:

We start by substituting x(t)=2 cos(t) and y(t)=4 sin(t) in h(x,y)=4x2+y2:

$h(x,y)=4x^{2}+y^{2}= 4(2cos(t))^{2}+(4sin(t))^{2}\\h(x,y)=4(4cos^{2}(t))+(16sen^{2}(t))\\h(x,y)=16cos^{2}(t)+16sen^{2}(t)=16(sen^{2}(t)+cos^{2}(t))\\h(x,y)=16$

So, in the path that the hiker chose:

$\frac{\partial{h}}{\partial{t}}=0$

Chain rule:

We start differentiating h(x,y) using chain rule as follows:

$\frac{\partial{h}}{\partial{t}}= \frac{\partial{h}}{\partial{x}}\frac{\partial{x}}{\partial{t}}+\frac{\partial{h}}{\partial{y}}\frac{\partial{y}}{\partial{t}}$

Now, it´s easy to find all these derivatives:

$\frac{\partial{h}}{\partial{x}}=8x\\\frac{\partial{x}}{\partial{t}}=-2sin(t)\\\frac{\partial{h}}{\partial{y}}=2y\\\frac{\partial{y}}{\partial{t}}=4cos(t)$

Now we replace them in the chain rule, with the replacement x=2cos(t) and y=4sin(t) in the x,y that are left and we operate everything:

$\frac{\partial{h}}{\partial{t}}= 8x(-2sin(t))+2y(4cos(t)$

$\frac{\partial{h}}{\partial{t}}= 8(2cos(t))(-2sin(t))+2(4sin(t))(4cos(t)$

$\frac{\partial{h}}{\partial{t}}= -32cos(t)sin(t)+32sin(t)cos(t)$

$\frac{\partial{h}}{\partial{t}}= 0$

This will be our answer

6 0

3 years ago