All categories

2 years ago

What is 140 rounded to the nearest hundredth

Mathematics

1 answer:

muminat2 years ago

7 0

I mean, there is no decimal place, so you can’t determine to the nearest hundredth place.

For the hundreds place, it would round to 100, though.

You might be interested in

Can someone tell me how much is 0,5-⁶?

serg [7]

Answer:

add this and (5 × 5 × 5) × (5 × 5 × 5 × 5 × 5

Step-by-step explanation:

3 0

2 years ago

Help please i need help if i get this one wrong i need to start over

earnstyle [38]

Answer:

Step-by-step explanation:

Right/left by 3 units because in the equation, it added 3. Right because it is positive and Right/Left because this represents the change in the x-axis.

Up/Down by 5 units because in the equation, it added 5. Up because it is positive and Up/Down because this represnts the change in the y-axis.

6 0

3 years ago

Which property can be used to expand the expression -27x+7)

Komok [63]

Answer: I think distributive

Step-by-step explanation:

4 0

2 years ago

One thousand tonnes (1000 t, one t equals 10 cubed kg) of sand contains about a trillion (10 super 12) grains of sand. How many

Phantasy [73]

Answer with Step-by-step explanation:

Since 1 mole of sand will contain Avagadro's Number of sand particles (by definition of 1 mole)

Thus we have

1 mole of sand = $6.022\times 10^{23}$ sand particles

Thus in number of trillion sand particles we have no of trillion sand particles in 12 mole is

$\frac{6.022\times 10^{23}}{10^{12}}=6.022\times 10^{11}$

Now since it is given that mass of 1 trillion sand particles is 1000 tonnes Thus the mass of $6.022\times 10^{11}$ trillion sand particles is

$Mass=1000\times 6.022\times 10^{11}=6.022\times 10^{14}tonnes$

Part 2)

Since it is given that volume of 1 sand particle is $1.0mm^{3}$ thus the volume of 1 mole of sand is volume of $6.022\times 10^{23}$ sand particles

Thus volume of 1 mole is $V=1.00\times 10^{-18}km^{3}\times 6.022\times 10^{23}=6.022\times 10^{5}km^{3}$

Now since the Area of united states is $A=3.6\times 10^{6}mile^{2}=5.8\times 10^{6}km^{2}$

Thus the depth of the sand pile is

$Depth=\frac{Volume}{Area}=\frac{6.022\times 10^5}{5.8\times 10^6}=0.10387km=103.8meters$

4 0

2 years ago

HELPPPP !!! ILL GIVE U BRAINLIEST

Pepsi [2]

I believe the answer is D because those equation indicate an area on the negative side

8 0

2 years ago

Read 2 more answers