All categories

3 years ago

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT!

Mathematics

2 answers:

Natali5045456 [20]3 years ago

7 0

The correct answer would be 6

frez [133]3 years ago

7 0

6 is the answer cause 6 x6 x 6 = 216

You might be interested in

What is (0.54)⋅(3.8)=

ra1l [238]

Answer:

2.052

Step-by-step explanation:

I hope it helped lot sj

8 0

2 years ago

Read 2 more answers

I MARK THE BRAINLIEST! PLEASE TELL ME IF THIS IS RIGHT AND IF IT IS WRONG TO TELL ME THE ANSWER NO GUESSING I WILL REPORT YOU!!!

jok3333 [9.3K]

Answer: (2,6) is correct

3 0

3 years ago

Read 2 more answers

The factory produces 4 gallons of orange juice for every 24 gallons of apple juice but is use this ratio to answer the question

kkurt [141]

Answer:

For every 1 gallon of Orange Juice the factory makes 6 gallons of Apple Juice so id say the ratio is 1:6

Step-by-step explanation:

8 0

2 years ago

Expand $\left(x^2 \frac{1}{x}\right)^3$. (Write the terms with higher degree first, so for example an $x^2$ term would come befo

Ilya [14]

Answer:

$x^6+3x^3+3+\dfrac{1}{x^3}$

Step-by-step explanation:

We can use the pascal triangle to find the coefficients of the expasion of the cube. It holds that

$(x^2 + \dfrac{1}{x})^3=(x^2)^3+3(x^2)^2\dfrac{1}{x}+3x^2(\dfrac{1}{x})^2+(\dfrac{1}{x})^3\\\\=x^6+3x^4\dfrac{1}{x}+3x^2\dfrac{1}{x^2}+\dfrac{1}{x^3}\\\\=x^6+3x^3+3+\dfrac{1}{x^3}$

4 0

3 years ago

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago