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2 years ago

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If the area of one face of a cube is 64cm^2, what is the cube's surface area? Please enter your answer in numbers only, don't en

ter the units.

2 answers:

Dafna1 [17]2 years ago

6 0

Since the area of one side of a cube is 64, you can find the surface area by adding the rest of the 5 sides which is:
64+64+64+64+64+64=384

Alecsey [184]2 years ago

4 0

Answer:

4096

Step-by-step explanation:

i think that is the answer because are you have to multiple them

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10.In Star School, the ratio of students who bring

Rina8888 [55]

Answer:

There are 406 students in the school

Step-by-step explanation:

Let the number of students in the school be x

The total ratio is 11 + 3 = 14

The ratio that buys lunch is represented by 3

So, we have it that;

3/14 * x = 87

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x = (14 * 87)/3

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2 years ago

Compare the rules of two games.

OverLord2011 [107]

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Step-by-step explanation:

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2 years ago

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Alik [6]

Answer:

-a^2 - a + 12

Step-by-step explanation:

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6 0

3 years ago

Read 2 more answers

Urgent. Please show all work

myrzilka [38]

Answer:

$\displaystyle f'(x) = \frac{4}{x^2}$

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Differentiation

Derivatives
Derivative Notation

The definition of a derivative is the slope of the tangent line: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

<em /> $\displaystyle f(x) = -\frac{4}{x}$

<u>Step 2: Differentiate</u>

[Function] Substitute in <em>x</em>: $\displaystyle f(x + h) = -\frac{4}{x + h}$
Substitute in functions [Definition of a Derivative]: $\displaystyle f'(x) = \lim_{h \to 0} \frac{-\frac{4}{x + h} - \big( -\frac{4}{x} \big)}{h}$
Simplify: $\displaystyle f'(x) = \lim_{h \to 0} \frac{4}{x(x+ h)}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle f'(x) = \frac{4}{x(x+ 0)}$
Simplify: $\displaystyle f'(x) = \frac{4}{x^2}$

∴ the derivative of the given function will be equal to 4 divided by x².

---

Learn more about derivatives: brainly.com/question/25804880

Learn more about calculus: brainly.com/question/23558817

---

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

2 years ago