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Lady bird [3.3K]
3 years ago
14

What was the greatest challenge cities faced as a result of rapid industrialization in the 1800s?

Mathematics
1 answer:
eduard3 years ago
7 0
<span>The greatest challenge cities faced as a result of rapid industrialization in the 1800 was the dramatic increase in the number of the people that are moving into the cities. There was mass movement of the people from the rural regions into the cities where industries are located. This led to overcrowding in the cities.</span>
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7b+11&gt;9b-13 need to find b&gt;?
STALIN [3.7K]
7b+11 > 9b-13\ \ \ \ /-11\\\\7b+11-11 > 9b-13-11\\\\7b > 9b-24\ \ \ \ /-9b\\\\7b-9b > 9b-9b-24\\\\-2b > -24\ \ \ \ /:(-2) < 0\ then\ " > "\ change\ to\ "\ < "\\\\b < 12\\\\b\in(-\infty;\ 12)

8 0
3 years ago
Find the slope of the line through each pair of points
ddd [48]
The answer is 27/23. The way you listed the answers is confusing so I’m not giving a letter. Because [17+10]/[17+6]= 27/23
5 0
3 years ago
What is the solution of -8/2y-8=5/y+4-7y+8/y^2-16 <br> y = –4 <br> y = –2 <br> y = 4 <br> y = 6
maria [59]
(-8)/(2y-8)=(5/(y+4))-7y+(8/(y^2-16))
(-4)/(y-4)=(5/(y+4))-7y+(8/(y+4)(y-4))
((-4)(y+4))/((y+4)(y-4))=((5(y-4))/(y+4)(y-4))-(7y(y+4)(y-4))/(y+4)(y-4))+(8/(y+4)(y-4))
(-4(y+4))=(5(y-4))-(7y(y+4)(y-4))+8
-4y-16=5y-20-(7y(y^2-16))+8
-4y-16=5y-20-7y^3+112y+8
-4y-16=117y-7y^3-12
-4=(121-7y^2)(y)
None of these choices would be equal to -4

4 0
3 years ago
Read 2 more answers
Can someone help me​
sashaice [31]

Answer:

16.25

Step-by-step explanation:

6.5 + 5 = 32.5/2 = 16.25

8 0
3 years ago
HELP PLZ GUYS I BEG YOU 15 PTS ITS EASYYYYYYY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Aliun [14]

Answer:

1) The linear equation in point and slope form is y - 67 = -4 × (x - 14)

2) The variables are;

a) The number of candies available = y

b) The number of days Jennifer eats the candies =

c) The slope, m = -4

3) Jennifer received 123 pieces of candies on Halloween

Step-by-step explanation:

The given parameters are;

The number of candies Jennifer eats everyday = 4 pieces

The number of days for which Jennifer eats the daily 4 candies = 14

The number of candies left at the end of the 14th day = 67 candies

1) We note that the rate of decrease in the number of candies = 4 candies/day

Therefore, the slope of the linear equation is m = -4

The y-intercept = The initial amount of candies Jennifer has = c = 67 + 14× 4 = 123 candies

The linear equation in point and slope form is given as follows;

y - 67 = -4 × (x - 14)

2) The variables are;

a) The y-value represents the number of candies available on a specific day

b) The x value represents the number of days Jennifer eats the candies'

c) The slope = The rate of decrease in the number of candies per day = -4

3) The number of candies Jennifer receives on Halloween is given by the y-intercept of the straight line equation as follows;

y - 67 = -4 × (x - 14)

y - 67 = -4·x + 56

y = -4·x + 56 + 67 = -4·x + 123

y = -4·x + 123

Comparing the above equation, with the general form of the straight line equation, y = m·x + c, where, the constant term, c = The y-intercept, we have;

The y-intercept of the equation y = -4·x + 123 = 123 = The initial amount of candies Jennifer received on Halloween.

6 0
3 years ago
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