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3 years ago

Estimating quotient for greater dividend

Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

7 0

Yup it has to be a greater divedend if not answer would be completly wrong

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REAAAALLLLY NEED HELP

pashok25 [27]

If you think about it, the question is asking us to find the greatest common factor, or GCF, of the two numbers, 24 and 18.
First, find all of the factors of 24.
The factors are: 1, 2, 3, 4, 6, 8, 12, 24
Next, find the factors of 18.
The factors are: 1, 2, 3, 6, 9, 18
List out all of the factors that both of the numbers have.
The factors are: 1, 2, 3, 6
Whichever is the greatest of these numbers is the GCF.
The GCF is 6, so the greatest number of groups he can make and still be able to win is 6.
Hope this helps!

6 0

4 years ago

Read 2 more answers

Write the time 8:30 then write this time in two other ways using words

guajiro [1.7K]

Half past eight or thirty minutes to nine

8 0

3 years ago

Help me on this math promblem it’s also a missing and is effecting me

Vlad1618 [11]

Answer:

$20.42

Step-by-step explanation:

Given that:

Price of tablet computer = $255.25

Sales tax = 8%

Sales tax will be calculated as follows:

tax amount = 0.08 (255.25)

tax amount = $20.42

So Azim will spend $20.42 on the sales tax.

i hope it will help you!

4 0

4 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$