A bat named Fatima ate 1050 dragon flies on four consecutive nights.?

Question

Each night she ate 25 more thatn on the night before. How many did she eat each night? Solve this algebraically.

Answer 1

is fatima the B*** in your calss??
Anyways,
let say the first night she ate x number of flies
second=x+25
third= x+50
fourth= x+75

x+x+25+x+50+x+75=1050
4x+150=1050
4x = 900
x=225
plug in the number for x and you'll get number for each night

Answer 2

1050= x + (x+25) + (x+50) + (x+75)
1050=4x + 150
1050 -150 = 4x
900=4x
900/4 = x = 225
so she ate 225 the first night
225+25 = 250 the second
250+25 = 275 the third
275+25 = 300 the fourth
total =225+250+275+300=1050

Answer 3

Let x stand for the first night and x+25 x+50 and x+75 stand for the 2nd,3rd, and 4th nights.

4x+150=1050
4x+150-150=1050-150
4x=900
x=225 1st night
x+25=250 2nd night
x+50=275 3rd night
x+75=300 4th night

225+250+275+300=1050

Answer 4

If the number of dragon flies eaten on the first night was d, then on the second night it was d+25, on the third it was d+25+25 or d+50, and on the last night it was d+75. So:

d + (d+25) + (d+50) + (d+75) = 1050

Combine like terms and solve for d. Then use that to find the amount of dragon flies eaten on the other three nights.

Answer 5

We take the first day as x.
so the second day is x+25, third is x+2(25) and so on.

so:
1050= x + x +25 +x +50 +x + 75
1050= 4x + 150
4x= 1050-150
x= 900/4 = 225

so 225 on the first night and you just add 25 for the following days.

Answer 6

first night = (x+25)

second night= (x+50)

third night= (x+75)

fourth night= (x+100)

all togehter Fati ate 1050 flies, so we set the sum of these four evenings to 1050.

(x+25)+(x+50)+(x+75)+(x+100)=1050

combine like terms

4x+250=1050

solve for x

x=200

Answer 7

Yeast infection worse at night?

Answer 8

Who wants to start a diet consisting on eating nothing that was ever alive?

Answer 9

chessyet has it but that bat must have been constapated