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3 years ago

12

Ashley used 24 cm of tape to wrap 6 presents how many presents did ashley wrape with 44cm of tape

1 answer:

nekit [7.7K]3 years ago

4 0

Answer:

11 packages

Step-by-step explanation:

24/6=4

44/4=11

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3 years ago

Find the sum of difference

murzikaleks [220]

Evaluate question answers
1. 6x7 -3
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42 - 3
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Answer
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12 + 21 + 3
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3. 5 x (-3) +7 - 6
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-5 x3+7 - 6
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-15 + 7- 6
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-14

Find the sum or difference answers

1. Simplify or expand
(16m 4 - 8m + 3) + (-7m4 - m2 - 8)

16m4 - 8m + 3 - 7m4 - m2 - 8
V
9m4 - 8m - 5 - m2
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9m 4 - m2 - 8m - 5
Answer
9m 4 - m2 - 8m - 5

2. Simplify or expand
(3x2 - 5x + 9) + (2x2 + 9x -4)+ (7x- 11)
v
3x2 - 5x + 9+ 2x2+9x - 4+7x-11
V
5x2 + 11x - 6
Answer
5x2 + 11x - 6

3. Simplify or expand
(3y2 - y - 6) - (42 + 5y+ 3)
V
3y2 - y - 6 - 4y2 - 5y - 3
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-y2 - 6y - 9
Answer
-y2 - 6y - 9

4. -7 -2+2x2+ 9x
2 Simplify or expand
-7 - 2+2x2+9x
V
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8 0

2 years ago

6 1/4 ft = how many yards

puteri [66]

2.083333 yards are in 6 1/4 feet

6 0

3 years ago

Read 2 more answers

A line goes through the point (–4,–2) and has an x-intercept of –1.

Keith_Richards [23]

Answer:

y = - x - 6

Step-by-step explanation:

y = mx + b

m = - 1

y = - x + b

(- 4, - 2)

- 2 = - (-4) + b

-2 = 4 + b

b = - 6

y = - x - 6

5 0

3 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

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