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3 years ago

(62.1) x (3.2) = ?? Please help urgent!!!

Mathematics

1 answer:

Ivanshal [37]3 years ago

5 0

Answer:

198.72

Step-by-step explanation:

if u remove the brakctes and multiplay both of them by 10 and then multiply (621x32=19872 then u divide it by 100 and u get the answer 198.72

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Solve 5x – 1 = 3x – 15. Explain each step, and include the Properties of Equality that you used.

artcher [175]

5x – 1 = 3x – 15

Subtraction Property of Equality (Subtract 3x from both sides)

2x - 1 = -15

Addition Property of Equality (Add 1 to both sides)

2x = - 14

Division Property of Equality (Divide both sides by 2)

x = -7

7 0

3 years ago

PLEASE ANSWER ASAP! The difference between two-thirds of a number and 4 is -92.

sleet_krkn [62]

Answer:

Step-by-step explanation:

2/3 x - 4 = - 92

add 4 to both sides

2/3 x = - 88

multiply both sides by 3

2 x = - 264

divide both sides by 2

this is ur answer

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Hope this helps!

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3 years ago

Read 2 more answers

Add:7ab,8ab,-10ab,-3ab<br><br><br>

storchak [24]

Answer:

2ab

Step-by-step explanation:

7ab+8ab+(-10ab)+(-3ab)=

=15ab-13ab= 2ab

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4 years ago

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What is it plssss???

erastovalidia [21]

Answer:

First option, 4 because 1/6 + 1/6 + 1/6 +1/6 = 4/6

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3 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

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

5 0

3 years ago