Response: a) 
b) The area has grown by 0.75 square meters.
Detailed breakdown:
Let the room's length be l
Let the room's breadth be 'b'
The area of a rectangle is calculated using the formula:
Per the problem, the room's new area equals 175% of its old area.
Thus, a) The new area of the room is determined as
This indicates a 175% increase in the room's length, resulting in a 175% increase in area.
b) The increase in square meters for Megan's room area is
Answer:
The main aspects are
50ft³ = volume
0.5ft³/min = filling rate
Time= 100 minutes
Step-by-step breakdown:
Chiang is in the process of filling a 50ft container with water at a speed of 0.5 ft/min.
50ft³ = volume
0.5ft³/min = flow rate
It’s important to note that we still need to determine how long it will take to fill the 50 ft³ container.
Time = volume / rate
Time = 50 ft³ / 0.5 ft³/min
Time = 50 / 0.5 min
Time = 100 min
The main aspects are
50ft³ = volume
0.5ft³/min = filling rate
Time= 100 minutes
The outcome is "
GK=37, GH=14, and JK=19".The problem references a missing attachment file, thus please find it attached.
The provided values are:
GJ=4x+2
HK=6x-1
Given GK =37
Need to calculate:
HJ=?
From GK: GK = GJ+JK yields GJ+(HK-HJ)
= 4x+2+((6x-1)-x) yields
9x+1...(a)Substituting (a) into the equation gives: 9(4)+1 yields 37 solves to x=4... for answer (b) and (c). Substituting x=4 yields:
GH=14 and JK=19.
In this problem the number we are working with is:
105,159
By definition we note:
thousand place: a five-digit quantity greater than zero.
Moreover, the rounding rule is:
if the digit being removed is 5 or more, increase the kept digit by one.
Therefore, rounding to the nearest ten thousand yields:
105,159 = 110,000
Answer:
105,159 rounded to the nearest ten thousand is:
105,159 = 110,000
Answer:
Step-by-step explanation: The error made by the student was dividing the wins by the losses.
The student should have divided the wins by the total number of games played.
Initially, the student ought to have summed 20 and 10 to conclude there were 30 games in total.
