All categories

2 years ago

The length, height, and width of a cube are multiplied by 6. What effect does this have on the surface area of the cube?

Mathematics

1 answer:

Harman [31]2 years ago

4 0

Answer:

Surface Area increases by 36 times.

Step-by-step explanation:

$Let \ length, \ height \ and \ width \ of \ a \ cube \ be \ a$

$Surface \ Area = 6a^2$

$Now \ length, \ height \ and \ width \ are \ multiplied \ by \ 6, that \ is \ sides = 6a$

$New \ Surface\ area = 6 (6a)^2 = 6 (36a^2) = 36 \times 6a^2$

$That \ is \ new \ surface \ area = 36 \times \ old \ surface\ area$

You might be interested in

PLEASE HELP FOR 100 POINTS AND BRAINLIEST! Find the unknown measures. Round lengths to the nearest hundredth and angle measures

Varvara68 [4.7K]

Answer:

A. BC ≈ 2.52; AB ≈ 1.28; m∠C = 27°

B. BC ≈ 1.28; AB ≈ 2.52; m∠C = 27°

C. BC ≈ 2.52; AB ≈ 1.28; m∠C = 117°

D. BC ≈ 1.28; AB ≈ 2.52; m∠C = 117°

3 0

2 years ago

A company that teaches self-improvement seminars is holding one of its seminars in Lancaster. The company pays a flat fee of $98

alukav5142 [94]

Answer:

Step-by-step explanation:

Let x represent the number of attendees that it will take the company to break even.

The company pays a flat fee of $98 to rent a facility in which to hold each session. Additionally, for every attendee who registers, the company must spend $16 to purchase books and supplies. This means that the total cost that the company would pay for x attendees is

16x + 98

Each attendee will pay $65 for the seminar. This means that the total revenue that the company would generate from x attendees is 65x.

At the break even point,

total cost = total revenue

Therefore,

16x + 90 = 65x

65x - 16x = 98

49x = 98

x = 98/49

x = 2

It will take 2 attendees and the total expenses and revenues is

2 × 65 = $130

4 0

3 years ago

Determine the value of X for the drawing below.

ASHA 777 [7]

Answer: x = 1

Step-by-step explanation: In short, 0 is the only number such that for any number x, x + 0 = x. ... Well, it's the only number which can be multiplied by any other number without changing that other number. In short, the multiplicative identity is the number 1, because for any other number x, 1*x = x.

5 0

2 years ago

Read 2 more answers

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

2 years ago

Can I get help on this problem showing my work?

Evgen [1.6K]

$cost \: of \: 1 \:bracelet \\(no\:shipping): \\72 \: dollars - 9 \: dollars \: shipping = \\ 63 \: \div 9 \: = cost \: of \: 1 \: bracelet \: \\ = 7 \: dollars$
$\\cost \: of \: shipping: \\ 9 \: dollars \: \div 9 \: bracelets \: = \\ 1 \: dollar\: per\:bracelet \\ \\ formula \: for \: cost \: overall: \\ 7y \: + y$

y = quantity, for example:

I want to buy 1 bracelet =
1×7 = 7 + 1 = 8$ per bracelet

6 0

3 years ago

Read 2 more answers