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3 years ago

The equivalent expression for 4g+2 using distributive property

Mathematics

1 answer:

Mars2501 [29]3 years ago

5 0

Answer:

2(2g+1)

Explanation:

4g and 2 goes into 2

divide by 2

4g/2=2g

2/2=1

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Please answer this question.

Nastasia [14]

Answer:

2x - 5

Step-by-step explanation:

1/5(10x - 25)

Use the Distrubitve Property

1/5 * 10x = 2x

1/5 * 25 = 5

New equation is 2x - 5

4 0

3 years ago

What is the third quartile of this data set? 20, 21, 24, 25, 28, 29, 35, 37, 42

nata0808 [166]

Answer:

Step-by-step explanation:

⇒The question is on third quartile

⇒To find the third quartile we calculate the median of the upper half of the data

Arrange the data in an increasing order

20, 21, 24, 25, 28, 29, 35, 37, 42

Locate the median, the center value

20, 21, 24, 25, 28, 29, 35, 37, 42

The values 20, 21, 24, 25 ------------lower half used in finding first quartile Q1

The value 28 is the median

The vlaues 29, 35, 37, 42...............upper half used in finding 3rd quartile Q3

Finding third quartile Q3= median of the upper half

upper half= 29,35,37,42

median =( 35+37)/2 = 36

6 0

3 years ago

Solve equation W=a+2 z=a

MissTica

Answer:

Yyy

Step-by-step explanation:

4 0

3 years ago

A description of different houses on the market includes the following three variables. Which of these variables is quantitative

Grace [21]

Answer:

Step-by-step explanation:

D.All of the above

6 0

3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Maru [420]

Parameterize $S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$ .

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}$

6 0

3 years ago