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avanturin [10]
3 years ago
7

Answer all questionsif you cant figure out what a number is, just guess or skip it

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
8 0
2.4 ft/min goes with table 4
10 ft/min goes with table 2
-0.8 ft/min goes with table 1
0.25 ft/min goes with table 3
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Hello!
Lena [83]

9514 1404 393

Answer:

   square with side length √17 units, so area 17 square units.

Step-by-step explanation:

Each side has a slope with a rise of 1 and a run of 4, or a rise of -4 and a run of 1. The ratios of these numbers are opposite reciprocals, so we know the sides are perpendicular. The sum of squares of these numbers is 17, so we know the sides are √17 and all are the same length. (The Pythagorean theorem tells us this.)

A quadrilateral with equal-length perpendicular sides is a square. This square has side length √17 units and area (√17)² = 17 square units.

5 0
3 years ago
Mobistar is a mobile services company
dem82 [27]

Answer:

Yes it is a mobile services company

7 0
3 years ago
Please please please help me 19 points this is due today i know the answer is -47 but i dont know how to work the problem out to
Eduardwww [97]
So first we need to know this:

Two negative signs make a positive
Two positives make a positive
One positive and one negative makes a negative.

So we can turn the equation to -4 -44 - - 1
2 minuses make plus
-4 -44 +1 = -48 +1 = -47
5 0
3 years ago
PLEASEEEE SOMEONE HELPP WIRH THIS ANSWERER
mr_godi [17]
Y=X-5 is the answer
8 0
3 years ago
Read 2 more answers
A printer need to make a poster that will have a total poster area of 200 in^2 and will have 1 inch margins on the sides, a 2-in
Katyanochek1 [597]

Answer:

The dimensions that will give the largest printed area of 132.1669 in^2 are

= Length x Width

= 15.20829 * 8.69045

Step-by-step explanation:

a) Data and Calculations:

Total poster area = 200 in^2

Side margins = 1 inch each

Top and bottom margins = 2 inches each

Let x = length of the full poster

then 200/x = width of the full poster

Therefore, the length of the printed area = x - 3.5

and the width of the printed area = (200/x)-2

Therefore, the Area of the Printed space = (x-3.5)((200/x)-2)

Solving for the Area (A) of the printed space, we have

 

A = (x-3.5)(-200/x2) + ((200/x)-2)

 

A = 700 -2x2

 

If the derivative is set to 0, we have:

0 = 700 -2x2

 

700 = 2x2

 

350 = x2

 

x = 18.70829  The original length

 

width = 10.69045

 

Therefore, the area of space available for printing is

 

15.20829  *  8.69045

 

= 132.1669 in^2

5 0
3 years ago
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