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3 years ago

How does the value of the digit 3 in the number 63,297 compared to the value of the digit 3 in the number 60,325?

Mathematics

1 answer:

Elina [12.6K]3 years ago

4 0

Answer:

The value of digit 3 in 63,297 is in the thousands place while the digit 3 in the number 60,325 is in the hundreds place

Hope this helps!

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Each students in class 38 students was asked to report how many siblings he has. Data a 0/38 Data b 1/38 Data c 13/38 Data 8/38

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Answer:

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Step-by-step explanation:

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3 years ago

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A checking account is overdrawn if it has a negative balance. Marc’s account is overdrawn by $63. What will the balance be after

miss Akunina [59]

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4 years ago

A number, x, rounded to 1 decimal place is 3.7<br> Write down the error interval for x.

soldier1979 [14.2K]

Answer: (3.65, 3.74)

Step-by-step explanation:

Now, when we round a number to a given decimal place, we need to look at the decimal to the left of the decimal place where we are rounding.

if it is 5 or more, we round up

if it is smaller than 5, we round down.

Then for example, if we want to round:

4.534

to the first decimal place, we need to look at the second decimal place.

we can see that it is a 2.

Then we round down, and get:

4.5

Now we know that, when rounding to the first decimal place, we have:

x = 3.7

Then the maximum value that x could have is:

x = 3.74 (because here we round down to 3.7)

and the minimum value that x could have is:

x = 3.65 (because here we round up to 3.7)

Then the error interval for x is:

(3.65, 3.74)

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3 years ago

A 15 ft. telephone pole has a wire that extends from the top of the pole to the ground. The wire and the ground form a 42° angle

Lorico [155]

Answer:

Correct choice is A

Step-by-step explanation:

Consider right triangle ABC formed by the telephone pole (side AB) and ground (side BC). In this triangle AC is the length of the wire, BC is the distance from the base of the pole to the spot where the wire touches the ground and AB=15 ft, ∠ACB=42°.

Then

$\sin 42^{\circ}=\dfrac{AB}{AC}\Rightarrow AC=\dfrac{AB}{\sin 42^{\circ}}=\dfrac{15}{\sin 42^{\circ}}\approx 22.4\ ft.$
$\tan 42^{\circ}=\dfrac{AB}{BC}\Rightarrow BC=\dfrac{AB}{\tan 42^{\circ}}=\dfrac{15}{\sin 42^{\circ}}\approx 16.7\ ft.$

8 0

3 years ago