All categories

3 years ago

Show that 4128 0012 4389 0110 is an invalid VISA number card

1 answer:

7 0

visa cards begin with 4 and then have 13 or 16 digits but here we have a number beginning with 4 and then it has the 15 digits.

<u>Step-by-step explanation:</u>

Each card has a different first digit number. The maximum length of this type of credit card number can be up to 19 digits.
In the visa card starting 6 digits are always said to be the issuer identifier.
The visa cards will have the starting digits as 4 whereas in Mastercard the starting digit is 5 but it has 16 digits numbers.

You might be interested in

I need help please and thank you

ExtremeBDS [4]

Answer:

D=340

Step-by-step explanation:

360-180/9

360-20

340

4 0

3 years ago

HELP PLEASE ANSWER QUICKLY!

quester [9]

Answer:

3 twelths is your answer :)

8 0

3 years ago

Read 2 more answers

Simplify completely 8x^3 ÷ 6x^2 - 4x / -2x

artcher [175]

Answer:

Simplified form is $-(\frac{8x^{3}}{3x-2})$ .

Step-by-step explanation:

The given expression is $8x^{3}\div \frac{6x^{2}-4x}{-2x}$

We can rewrite the given expression as $= 8x^{3}\times (-\frac{2x}{{6x^{2}-4x}})$

Further simplification of the expression gives $=\frac{-16x^{4}}{6x^{2}-4x}$

$=\frac{-16x^{4}}{2x(3x-2)}$

$=-(\frac{8x^{3}}{3x-2})$

So the simplified of the given expression is $=[tex]-(\frac{8x^{3}}{3x-2})$

3 0

3 years ago

Read 2 more answers

Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices

Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

$\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt$

Where P, Q are scalar functions

We want to compute

$\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy$

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths $\large C_1,C_2,C_3,C_4$ which represents the sides of the rectangle.

$\large C_1$ is the line segment from (0,0) to (5,0)

$\large C_2$ is the line segment from (5,0) to (5,1)

$\large C_3$ is the line segment from (5,1) to (0,1)

$\large C_4$ is the line segment from (0,1) to (0,0)

Then

$\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}$

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize $\large C_1$ as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

$\large \displaystyle\int_{C_1}xydx+x^2dy=0$

Now the second integral. We parametrize $\large C_2$ as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

$\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25$

The third integral. We parametrize $\large C_3$ as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

$\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2$

The fourth integral. We parametrize $\large C_4$ as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

$\large \displaystyle\int_{C_4}xydx+x^2dy=0$

$\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2$

Now, let us compute the value using Green's theorem.

According with this theorem

$\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx$

where A is the interior of the rectangle.

so A={(x,y) | 0≤ x≤ 5, 0≤ y≤ 1}

We have

$\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x$

$\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2$

3 0

3 years ago

X4y(4) = yx2(13), calculate (x+y)2

Delicious77 [7]

Please, for clarity, use " ^ " to denote exponentiation:

Correct format: x^4*y*(4) = y*x^2*(13)

This is an educated guess regarding what you meant to share. Please err on the side of using more parentheses ( ) to show which math operations are to be done first.

Your (x+y)2, better written as (x+y)^2, equals x^2 + 2xy + y^2, when expanded.

The question here is whether you can find this x^2 + 2xy + y^2 in your

"X4y(4) = yx2(13)"

Please lend a hand here. If at all possible obtain an image of the original version of this problem and share it. That's the only way to ensure that your helpers won't have to guess what the problem actually looks like.

3 0

3 years ago