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3 years ago

HURRRY PLEASE ANSWER I ONLY HAVE 5 MINN which expressions are equivalent to - 72 + (-6.5) + 2.5z + 4y - 1.5

Mathematics

1 answer:

STALIN [3.7K]3 years ago

8 0

Answer:

fghwjklpkjqkhb cghdbhn bwn x wxhw hwbmx

Step-by-step explanation:

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(who are you riddle) I'm smaller than 24. I am a common multiple of 7 and 3

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Charra [1.4K]

Answer:

137.5 Miles

Step-by-step explanation:

2 = 50 Right? So if you were to divide 50 by 2, you would get 1 = 25. All you would have to do from there would be to multiply 25 times 5.5 to get 137.5.

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4 years ago

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Please help me! I have no idea how to do this

elena-s [515]

The best way to do this is to make up values for the sides.

Say the lengths of square B's sides are each 4 cm. That means A's sides are 2 cm.

So, the area of square B is 4^2 = 16 cm^2, and A's area is 4 cm^2. We can see the shaded area is half of A, so it's 2 cm^2.

What percent of 16 is 2?

2 / 16 * 100 = 12.5%

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3 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