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3 years ago

When 8 is multiply by 2/3, the product is?

Mathematics

1 answer:

Korvikt [17]3 years ago

7 0

Answer:

8/3 or 5 1/3

Step-by-step explanation:

here's a visual tutorial I don't feel like writing, sorry

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Someone plz help me with this plz

kolezko [41]

Answer:

NOt finished

Step-by-step explanation:

PAGE 1

1.) enlargement b/c the number is more that 1

2.) reduction b/c the number is less than 1

3.) enlargement b/c the number is more than 1

4.) reduction b/c the number is less than 1

5.) reduction k = 1/3

6.) enlargement k = 5/2 or 2.5

PAGE 2

7.) reduction k = 1/3

8.) enlargement k = 2

PAGE 3

Screen shot

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screen shot

PAGE 5

screen shot

I WILL DO THE SCREENSHOTS TOMORROW

7 0

3 years ago

What is the solution to the system of equations graphed below

aliina [53]

The point where the lines intersect is the point that satisfies both equations.

It is (1, 5), selection C.

3 0

3 years ago

Read 2 more answers

Yo make lemonade you have to mix 2 quarts is water with 8 tablespoons is lemonade mix. How much mix will you need for 12 quarts

Alisiya [41]

2 quarts- 8 tablespoons
12 quarts- 8 x 6 = 48 tablespoons

6 0

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

Find AB distance. Please help

Svet_ta [14]

Answer:

Solution,

Let,(x1,y1)=(-4,-3)

(x2,y2)=(3,5)

Using distance formula,

AB=√(x2-x1)²+(y2-y1)²

=√(3+4)²+(5+3)²

=√49+64

=√113 units.

Step-by-step explanation:

7 0

2 years ago

When 8 is multiply by 2/3, the product is? ​

When 8 is multiply by 2/3, the product is?