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Serggg [28]
4 years ago
11

Divide m2n2/p3 by mp/n2 .

Mathematics
2 answers:
PilotLPTM [1.2K]4 years ago
4 0

Answer:

The answer is 'm'.

Step-by-step explanation:

Here, the given expression,

\frac{\frac{m^2n^2}{p^3}}{\frac{mp}{n^2}}

=\frac{m^2n^2\times n^2}{p^3\times mp}  

=\frac{m^2n^{2+2}}{p^{3+1}m}   ( a^m.a^n=a^{m+n} )

=\frac{m^2p^4}{p^4m}

=m^2p^4p^{-4}m^{-1}    ( a^{m}=\frac{1}{a^{-m}} )

=m^{2-1}p^{4-4}

=m^1p^0

=m

mr Goodwill [35]4 years ago
3 0
8n^2  /  3p^2

Hope it helped!
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Use two colors of counters to show a way to make 5.color to show the counters . Write the numbers to show the pair that makes 5
Fantom [35]

Answer:

1 and 4, 2 and 3, and 5 and 0.

Step-by-step explanation:

Counters are used by educators to teach children how to do calculations like addition, subtraction, multiplication and division. Using color counters help to visualise the result from the operands in a calculation.

To get the pairs of numbers that adds to 5.

- Use the first color counter to mark out the five ones on the sheet.

- then with the second color counter, subtract one from the five ones.

- repeat the first two steps and increment the subtracted value until it gets to five.

4 0
4 years ago
Determine the domain of the function.
AnnyKZ [126]
It it is the first choice i mentioned then B is the answer.
7 0
4 years ago
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Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)
sasho [114]

Answer:

The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ABC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|

On solving we get:

Area\,of\,triangle\,=25.

In Δ ADC:

A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ADC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|

On solving it by same process as above we get:

Area\,of\,triangle\,=114

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

25+114\\=25+114\\=139\ unit^2

Therefore, area of quadrilateral ABCD is 139 unit^2.

4 0
3 years ago
Draw 10 marbles use 5-group.
LenaWriter [7]

Answer:

2

Step-by-step explanation:

there would be 2 in every group, 10 divided by 5

4 0
3 years ago
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If the inverse of a square root function is a quadratic function, why do the graphs differ
Travka [436]
They don’t. The square root function is a horizontal quadratic.
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