All categories

2 years ago

How do I round 0.3 to the nearest thousandth

1 answer:

6 0

0.3 is 3 tenths.
The order of "ths" is tenths, hundredths, thousandths.
Since there are no numbers after the 3, you simply have to add 2 zeros to make it 300 thousandths.

0.300

You might be interested in

Which value of x makes the equation -5x-(-7-4x) =-2(3x-4 true?

Ludmilka [50]

Answer:

x = 1/5

Step-by-step explanation:

Step 1: Write equation

-5x - (-7 - 4x) = -2(3x - 4)

Step 2: Solve for x

Distribute: -5x + 7 + 4x = -6x + 8
Combine like terms: -x + 7 = -6x + 8
Add 6x to both sides: 5x + 7 = 8
Subtract 7 on both sides: 5x = 1
Divide by 5 on both sides: x = 1/5

Step 3: Check

Plug in x to verify it's a solution.

-5(1/5) - (-7 - 4(1/5)) = -2(3(1/5) - 4)

-1 - (-7 - 4/5) = -2(3/5 - 4)

-1 - (-39/5) = -2(-17/5)

-1 + 39/5 = 34/5

34/5 = 34/5

5 0

3 years ago

emily went to see a film at the local cinema.it started at quarter past two in the afternoon.it took her 25 minutes to walk from

Anastasy [175]

A quarter past two means 2:15. So if it took her 25 minutes to get there, the latest she could leave would be 1:50.

8 0

2 years ago

Read 2 more answers

Change at least one number in the table so that it DOES represent a function.

almond37 [142]

Answer:

change 2 to 4

and 1 to 5

1 --> 5

2 --> 10

3 --> 15

4 --> 20

5 --> 24

Step-by-step explanation:

Function means you can have one input for evry one output

so you have to change the 2 and 1 to diffrent numbers like 4 and 5

so now evey imput has their own out put

befor input one it had 5 and 24

and input two had 10 and 20 but changing the number you get one for one

6 0

2 years ago

Which shows the correct substitution of the values a, b, and c from the equation –2 = –x + x2 – 4 into the quadratic formula?

Serhud [2]

X=[-(-1)+or-sqrt ((-1)^2-4x1x-2)]/(2x1)

7 0

3 years ago

Read 2 more answers

Prove: Every point of S=(0, 1) is an interior point of S.

myrzilka [38]

Answer:

Proved

Step-by-step explanation:

To prove that every point in the open interval (0,1) is an interior point of S

This we can prove by contradiction method.

Let, if possible c be a point in the interval which is not an interior point.

Then c has a neighbourhood which contains atleast one point not in (0,1)

Let d be the point which is in neighbourhood of c but not in S(0,1)

Then the points between c and d would be either in (0,1) or not in (0,1)

If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.

Then we find that dn is a boundary point of S

But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.

Every point of S=(0, 1) is an interior point of S.

3 0

2 years ago