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4 years ago

Plz help I’ll give 60 I REALLY need help

Mathematics

2 answers:

boyakko [2]4 years ago

4 0

Answer:

a for the first one, i pretty sure

White raven [17]4 years ago

3 0

C for the secondo one :)

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A cryptographic hash takes a message as input and produces a fixed-length string as output, called the digital fingerprint. A br

Allushta [10]

Answer:

$1.85x10^{19}$ attempts are required to find a matching pair if the digital fingerprint is 64 bits long.

$3.40*10^{38}$ attempts are required to find a matching pair if the digital fingerprint is 128 bits long.

Step-by-step explanation:

Each bit has two options. So

How many attempts are required to find a matching pair if the digital fingerprint is 64 bits long?

So for each of the 64 bits, we have the following number of options.

2 - 2 - 2 - 2 -... - 2

So, in all, there are

$T = 2^{64} = 1.85x10^{19}$

options.

So, $1.85x10^{19}$ attempts are required to find a matching pair if the digital fingerprint is 64 bits long.

128 bits long?

Using the same logic as the first question.

$T = 2^{128} = 3.40*10^{38}$

So, $3.40*10^{38}$ attempts are required to find a matching pair if the digital fingerprint is 128 bits long.

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3 years ago

A computer sales associate makes $110 each day that she works and makes approximately $19 in commission for each computer that s

ioda

The computer salesman would have to sell 12 computers.

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3 years ago

At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not

Nikitich [7]

If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

Given the paraboloid

$y = x^2 + z^2$ .

We can rewrite it as a scalar value function f as follows:

$f(x,y,z)=x^2-y+z^2=0$

The normal to the paraboloid at any point is given by:

$\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\ \\ =2xi-j+2zk$

Also, the normal to the given plane $3x + 2y + 7z = 2$ is given by:

$3i+2j+7k$

Equating the two normal vectors, we have:

$2x=3\Rightarrow x= \frac{3}{2} \\ \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}$

Since, -1 = 2 is not possible, therefore there exist no such point on the paraboloid $y = x^2 + z^2$ such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2.

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the quota rent per pound is: $17.50.

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2 years ago

If Sam had 1099 cookies and she gave gen. 1000 how many cookies would Sam have left

TiliK225 [7]

He would have 99 cookies left

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3 years ago

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