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mel-nik [20]
3 years ago
6

How many thousands are in 5,000 ones

Mathematics
2 answers:
Anon25 [30]3 years ago
8 0
Should be 5. im not sure tho because i haven’t done this since fifth grade.
Iteru [2.4K]3 years ago
3 0
5 thousands in 5,000 ones.
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Three consecutive integers have a sum of 168. What is the value of the largest integer
Svet_ta [14]
Largest integer: 57

x + x + 1 + x + 2 = 168 
3x + 3 = 168
-3             -3
3x       = 165
/3             /3 
x         = 55 

x = smallest integer = 55
x + 1 = middle integer = 55 + 1 = 56
x + 2 = largest integer = 55 + 2 = 57 
6 0
3 years ago
I need help ASAP!!Explain the answer
nordsb [41]

Answer:

volume=pi times r^2 height÷3.

r is the radius, or in this case 4yd. height is 10yd. pi is 3.14

Here are the steps:

3.14r^2 height÷3

3.14×4^2 10÷3

=167.55

final answer rounded to the nearest tenth: 167.6

I apologize for getting the math wrong the first time. My formula was correct, but I needed to double check my calculations. I'm certain this is the right answer.

5 0
3 years ago
Uptown Photography charges $15 per portrait sheet, plus $19 for a session fee. Downtown Photography charges $29 for a session fe
stepan [7]

The answer is (2,49).


7 0
3 years ago
Eloise made a list of some multiples of 8/5 write 5 fractions that can be in Eloise list
alexgriva [62]
This question is asking for a list of fractions, that when you pull out common factors, they will all simplify back to the 8/5 fraction.

So if 2 is the common factor, you multiply the numerator and denominator by 2.

2: (8*2)/(5*2)= 16/10

3: (8*3)/5*3)= 24/15

4: (8*4)/(5*4)= 32/20

5: (8*5)/(5*5)= 40/25

10: (8*10)/5*10)= 80/50

Eloise's Potential List:
16/10, 24/15, 32/20, 40/25, 80/50

If you simplify any of these fractions above, you will get 8/5 again.

To work backwards:
80/50: 10 goes into 80, 10 goes into 50= 8/5

16/10: 2 goes into 16, 2 goes into 10= 8/5

Hope this helps! :)
4 0
3 years ago
If x – 10 is a factor of x2 – 8x – 20, what is the other factor?
Svetach [21]

Answer:

The other factor is (x+2)

Step-by-step explanation:

we know that

(x-a)(x-b)=x^{2}-xb-xa+ab

(x-a)(x-b)=x^{2}-(a+b)x+ab

In this problem we have

x^{2} -8x-20

and

a=10 -----> because is a factor

substitute and solve for b

x^{2} -8x-20=x^{2}-(10+b)x+10b

so

8=10+b\\b=-2

Verify in the second equation

-20=10b -----> -20=10(-2) -----> -20=-20--> is ok

The other factor is (x+2)

3 0
3 years ago
Read 2 more answers
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