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3 years ago

70,000 devided by 7,000

1 answer:

7 0

It would be 10 because 7000 can go into 70,000 10 times. You can you a calculator or look at the zeros take away the 3 zeros in 7000 and then 3 zeros in 70,000 making it 7 and 70 and 7 goes into 70 10 times as well. I hope this helps!

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Solve the equation 1/2x+1/4x= 16 A. 16 B. 21.3 C. 32 D. 64

Otrada [13]

1/2+1/4= 2/4+1/4= 3/4 or .75
Then,
3/4x=16
(3/4x)4/3=(16)4/3 *multiply by the reciprocal everytime you want to isolate x with a fraction\

x= 12

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2 years ago

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Step-by-step explanation:

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3 years ago

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Burka [1]

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Step-by-step explanation:

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2 years ago

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.

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3 years ago