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gogolik [260]
3 years ago
10

Help my sister with this pls

Mathematics
1 answer:
NeX [460]3 years ago
3 0

Answer:

23

Step-by-step explanation:

i think

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Did you ever free club is selling wrapping paper for a fundraiser they order 16 cases if each student in the club takes two thir
omeli [17]
16=2/3 times number of students
times both sides by 3
48=2 times number of students
divide both sides by 2
24=number of students

there aer 24 students i the club
3 0
3 years ago
Read 2 more answers
1. m – 7 < 6 A) m < –1 B) m > 1 C) m < 13 <---------------------- D) m < –13 2. x + 4.5 ≥ 5.5 A) x ≥ 10 B) x ≥
ikadub [295]
1) m - 7 < 6
 Add 7 to the both sides, 
m - 7 + 7 < 6 + 7
m < 13

So, Option C is your answer for this part.

2) x+4.5 ≥ 5.5
Subtract 4.5 from both sides, 
 x ≥ 1

So, Option B is your answer for this part.

3) p +12 > 9
Subtract 12 from the both sides, 
p > -3

So, Option D is your answer for this part.

4) x - 5 ≥ 9
So, Option A is your answer for this part

Hope this helps!

6 0
3 years ago
A circle with a radius of 15 ft is subdivided by a central angle. If the resulting arc length of the subtended arc is 10pi feet,
butalik [34]

Answer:

that is the answer to the question

4 0
3 years ago
I need to know what number 9 is I don’t understand it
Taya2010 [7]

The answer is (6/25). I just multiplied the 2 numbers, but I'll explain with something simpler:

Take the number 100, let's say I want (1/10) from that 100,

and from that (1/10), I want (4/5).

Well a tenth of 100 is 10, and four fifths of 10 is 8, thus we have 8/100 as a total (or 4/50)

The same answer can be found when multiplying (1/10) and (4/5), giving us (4/50) or (8/100)

7 0
3 years ago
The Population of a town was
aliina [53]

Answer:

33.3%

Step-by-step explanation:

Step one:

given data

Initial population (1980)= 12000

Final population(1990)= 16000

change in population= 16000-12000= 4000

Step two:

% increase= change in population/initail population *100

%increase= 4000/12000*100

%increase= 0.333*100

% increase= 33.3%

7 0
3 years ago
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