Please can you help me with my question! I'm desperate! It's due for Monday:

1 answer:

6 0

I observed that if I remove one (or all 8) piece(s) in the corners, and only them without adjacent ones, the total area does not change.

I consider the surface area of a small square as a unit of surface.

First class of solutions:

I removed all eight corners, leaving the total area unchanged.

I removed the central cube of the top surface obtaining an increase of the surface area with four units.

I removed one cube from the middle of an edge at the top (any of the four remaining) and I arrived at a figure with ten cubes less then the original one but with the same surface area.

(There's a lot more solutions here: https://nrich.maths.org/787/solution)

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Whats mc mean in mc2

Grace [21]

Answer:

the equation says that energy and mass (matter) are interchangeable; they are different forms of the same thing.

Step-by-step explanation:

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2 years ago

reposting cuz i need this asap for the one i circled do i find the hypotenuse or leg? pls help. i don’t need the answer i just n

Kaylis [27]

Answer:

Hypotenuse

Step-by-step explanation:

For this problem you are given the vertical and horizontal sides of the triangle. The flagpole (6.5 meters) is the up-and-down triangle side and the distance from the hook to the flagpole is the base (5.2 meters) of the triangle.

Hope this helps!

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2 years ago

The denominator of a fraction is 9 more than the numerator. if both the numerator and the denominator are decreased by 1 the res

yarga [219]

X/x+9 (original fraction)

x+1/ x+10 = 4/13

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now that you have the value of x, substitute it into the original fraction to get:

3/12

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2 years ago

WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$