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Hitman42 [59]
3 years ago
9

Describe how the graph of y= x2 can be transformed to the graph of the given equation. (2 points)

Mathematics
1 answer:
Darya [45]3 years ago
5 0
1. Is too easy it’s 300
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Can some break this down step by step please I'm confused?​
frutty [35]

take 16+5=21. 29-21=8 so answer is 8

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4 years ago
2x + 7= -11 please help
Dvinal [7]

Answer:

x = -9

Step-by-step explanation:

You are solving for x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

First, subtract 7 from both sides:

2x + 7 (-7) = -11 (-7)

2x = -11 - 7

2x = -18

Next, divide 2 from both sides:

(2x)/2 = (-18)/2

x = -18/2

x = -9

x = -9 is your answer.

~

6 0
3 years ago
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Simplify. 2,500,000 to the zero power
Sophie [7]

Answer:

1

Step-by-step explanation:

any number that is raised to the zero power will equal to 1. hope this helps!

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What is the slope of the line? <br> A. -1<br> B. 0<br> C. 1 <br> D. 20
sp2606 [1]
The answer is B) 100%
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3 years ago
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URGENT please help!!! I need a explanation please!! You own a restaurant and can reopen as long as people seated at a table are
hram777 [196]

Answer:

1) Maximum number of tables that can be kept in the dinning room is 10 tables

2)The maximum number of customers, given 6 people per table is 60 customers

Step-by-step explanation:

The parameters given are;

Distance between people seated at different tables = 6 ft

Dimension of table = 2 m by 1.5 m

Number of people at a table = 6 people

Dimension of dinning;

Width = 12 m

Length = 20 m

Dimension of entrance;

Width = 20/21*30  m

Length = 10 m

With 6 meters between tables and a table with of 1.5, we have for the arrangement in the question;

3 tables = 4.5 m

Distance between = 2 × 6 = 12

4.5 + 12 = 16 m (More room required)

With two tables, we have;

Width of tables = 2×1.5 = 3 m

Distance between = 6 m

Dining room width required = 3 + 6 = 9 m

Therefore, the maximum tables in each row = 2

Given that the dining room area extends to the entrance area, total length = 30 m.

With 4 tables we have;

Length = 4 × 2 + 6 × 3 = 26

While on the entrance side of the dinning room area which is 20 m, we have 3 tables;

Length = 3 × 2 + 6 × 2 = 18

Therefore, both arrangements are acceptable;

Just on entering by the right the length = 20/21×30 m

Therefore, with 3 tables will required number due to proximity wit the door and tables arranged along the right wall of the dinning

1) Maximum number of tables that can be kept in the dinning room = 4 + 3 + 3 = 10 tables

2)The maximum number of customers, given 6 people per table = 6×10 = 60 customers.

4 0
3 years ago
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