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3 years ago

72 games 20% discount

2 answers:

5 0

The answer is 14.4 because, if multiply 72 by .2 you would get 14.4. 20% is also like .5, or .05. It actually depends !!! Hope this helps

Firdavs [7]3 years ago

3 0

72*.2=14.4

72-14.4= 57.6

Games cost $57.60

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An arena is hosting a concert. At the most, the arena can hold 8,500 people. If tickets have already been sold to 6,900 people,

ElenaW [278]

Answer:

Subtract 6,900 from 8,500 to get the highest number of tickets that can still be sold which is 1,600.

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3 years ago

What is the remainder when 3 is synthetically divided into the polynomial -2x^2 + 7x - 9?

goblinko [34]

Answer:

for apex the answer is -6

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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Blababa [14]

|
|__x_- 4___________ 2x³/2x² =x
2x² + 2x +3 | 2x³ - 6x² +7x +3
- (<span>2x³ +2x² +3x)
</span> -8x² +4x +3 -8x²/2x² = -4
<span>-(-8x² -8x -12)
</span> 12x +15

(x-4) + (12x +15)/(2x² + 2x +3)

6 0

3 years ago

1. Ima Thief robs a bank and takes off in a getaway car. Five minutes Polly Srule takes off after Ima.

chubhunter [2.5K]

Answer:

a. 12 minutes

b. 34 minutes

Step-by-step explanation:

Here, we are told that Ima has driven 5 minutes before Polly started driving

So if Ima has driven for x minutes , polly would have driven for y minutes but the difference between x and y is 5

So mathematically;

x = y + 5

a. 17 = y + 5

y = 17-5 = 12 minutes

b. x = 29 + 5

x = 34 minutes

8 0

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.

5 0

3 years ago