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3 years ago

Write a formula for the sequence of cube numbers: 1, 8, 27, 64, ….

Mathematics

1 answer:

Marianna [84]3 years ago

4 0

Answer: $n^3$ , where n = 1,2,3,4,.....

Step-by-step explanation:

The given sequence : 1, 8, 27, 64, ….

We observe that the terms are perfect cubes of consecutive integers (starting from 1) .

Terms can be written as:

First term = $a_1=1=1^3$

Second term= $a_2=8=2^3$

Third term = $a_3=27=3^3$

Fourth term = $a_4=64=4^3$

Thus , for any nth term ,the formula for the sequence is : $n^3$ , where n = 1,2,3,4,.....

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12 1/2 cups is equavelent

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3 years ago

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Answer:

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3 years ago

What is the standard form of 3^5

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3 years ago

Your manufacturing company sells metal nuts and bolts that come together in a package. The diameter of the bolts coming off the

Reil [10]

Answer:

The z score for bolt of diameter 18.12 mm is 1.20.

Step-by-step explanation:

Let X = diameter of bolts.

It is provided that the random variable X follows a Normal distribution with mean, μ = 18 mm and standard deviation, σ = 0.10 mm.

A z-score is a standardized score, a numerical, that defines how far a data value from the mean.

The distribution of z-scores is defined by the Standard Normal distribution.

$Z\sim N(0, 1)$

The formula to compute the z-score is:

$z=\frac{x-\mu}{\sigma}$

The value of the diameter of a bolt is, x = 18.12 mm.

Compute the z-score for this value as follows:

$z=\frac{x-\mu}{\sigma}=\frac{18.12-18}{0.10}=\frac{0.12}{0.10}=1.20$

Thus, the z score for bolt of diameter 18.12 mm is 1.20.

5 0

3 years ago

Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .

7 0

3 years ago