A cylinder has a base radius of 3 centimeters and a height of 15 centimeters.

Z is proportional to x^3. describe exactly what will happen to x when z became 8 times greater

Reil [10]

Answer: x is doubled.

Step-by-step explanation:

A proportional relationship between x and y is written as:

y = k*x

where k is called the constant of proportionality.

In this case, we know that we have a proportional relationship between z and x^3

Then:

z = k*x^3

What will happen to x when z is 8 times greater?

Let's rewrite this equation for two new quantities, z' and x'

z' = k*(x')^3

Now we need to replace z' by 8*z, then:

8*z = k*(x')^3

We want to find a relationship between x' and x.

And by the first relationship, we know that:

z = k*x^3

Then we can replace this in the equation "8*z = k*(x')^3" to get:

8*(k*x^3) = k*(x')^3

8*k*x^3 = k*(x')^3

Now we can divide both sides by k, so we get:

8*x^3 = (x')^3

Now we can apply the cubic root to both sides, to get:

∛(8*x^3) = ∛(x')^3

∛(8)*x = x'

2*x = x'

Then when we increase the value of z 8 times, the value of x will be doubled.

8 0

2 years ago

The sign of the product of -35 and -625 is positive, negative, or zero

IRISSAK [1]

Answer:

Positive

Step-by-step explanation:

The product of two negative numbers has a positive sign, whereas the product of a positive and a negative number is negative.

Since -35 and -625 are both negative, they would have a positive sign for their product.

Hope this helps.

7 0

3 years ago

Estimating a square root.

Digiron [165]

I'm sorry but I'm not into square roots yet :(

5 0

3 years ago

Read 2 more answers

Consider the set {4, 5, 7, 7, 8, 8, 12, 13, 13, 15, 18, 20, 22, 24, 26, 27, 27, 37, 43}. What would be an appropriate interval t

adelina 88 [10]

Proper interval would be 5

8 0

3 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago