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3 years ago

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Mathematics

1 answer:

denis23 [38]3 years ago

6 0

Answer:

3y5432 -2123466

Step-by-step explanation:

that is the answer simplified but I don't know what you're asking for

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Evaluate the summation of negative 5 n minus 1, from n equals 3 to 12..

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The summation of negative 5 n minus 1, from n equals 3 to 12 can be expressed as

n = 3/12

substitute the value of n to the equatio

-5n – 1

-5( 3/12 ) -1

-9/4

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After a holiday dinner there are 3 1/3 Apple pies left and 2 5/6 pumpkin pie left .

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In all poe ate 2/3 of the left overs

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A 20° sector in a circle has an area of 21.51 pi yd?.

kodGreya [7K]

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This would be 21.5π * 360 / 20 yd^2 = 1215.18 yd^2 ( using pi = 3.14)

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3 years ago

What is the probability of rolling a dice and getting either a 1 or 2 and flipping a coin and getting a heads?

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago