Answer:
-2.7 < y
Step-by-step explanation:
2.9 < 5.6+y
Subtract 5.6 from each side
2.9-5.6 < 5.6-5.6+y
-2.7 < y
Answer: Hello your question some data but i will provide a general solution based on the scope of your question making some suggestions as well
answer : Summation of displacements ( back and forth distance ) / Number of Runners
Step-by-step explanation:
Given that ; the aim of the race is to raise money
The number of miles/ distance covered will determine how much money that would be raised
Formula to resolve the problem = Summation of displacements ( back and forth distance ) / Number of Runners
<em>Lets assume: ( example ) </em>
<em> Distance between the Park and the City hall is = 6 miles </em>
<em>Number of runners = 4</em>
<em>Given that the runners Run from the Park to the City hall and then run back</em>
<em>Total miles covered by each runner = ( 6 + 6 )/ 4 = 12/4 = 3 miles </em>
Please don't just copy and paste or click an answer! PLEASE READ! I spend 14 minutes and I at least want you to read it!
11.2 * 1.4 = 15.68
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Answer:
<h3>(4-4 = 0) - (4x5 =20) xDDddd</h3>
Answer:
a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle
Step-by-step explanation:
Let l = the original length of the original rectangle
Let w = the original width of the original rectangle
From the description of the problem, we can construct the following two equations
l=2*w (Equation #1)
(l+4)*(w+4)=l*w+88 (Equation #2)
Substitute equation #1 into equation #2
(2w+4)*(w+4)=(2w*w)+88
2w^2+4w+8w+16=2w^2+88
collect like terms on the same side of the equation
2w^2+2w^2 +12w+16-88=0
4w^2+12w-72=0
Since 4 is afactor of each term, divide both sides of the equation by 4
w^2+3w-18=0
The quadratic equation can be factored into (w+6)*(w-3)=0
Therefore w=-6 or w=3
w=-6 can be rejected because the length of a rectangle can't be negative so
w=3 and from equation #1 l=2*w=2*3=6
I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.