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Alexeev081 [22]
3 years ago
9

Round to the nearest tenth ​

Mathematics
1 answer:
Mandarinka [93]3 years ago
5 0

Answer:

Step-by-step explanation:

Say you wanted to round the number 838.274. Depending on which place value you'll round to, the final result will vary. Rounding 838.274:

Rounding to the nearest hundred is 800

Rounding to the nearest ten is 840

Rounding to the nearest one is 838

Rounding to the nearest tenth is 838.3

Rounding to the nearest hundredth is 838.27

Basic Rules of Rounding

When you "round to the nearest _____" regardless of what goes in the blank the steps are nearly always the same:

Identify which place value you are rounding to. The smaller the place value, the more accurate the final result will be.

Look to the next smallest place value, the digit to the right of the place value you're rounding to. For example, if you want to round to the nearest ten you'd look at the ones place.

If the digit in the next smallest place value is less than five (0, 1, 2, 3, or 4), you leave the digit you want to round to as-is. Any digits after that number (including the next smallest place value you just looked at) become zeros, or drop-off if they're located after the decimal point. This is called rounding down.

If the next smallest place value is greater than or equal to five (5, 6, 7, 8, or 9), you increase the value of the digit you're rounding to by one (+1). Just like before, any remaining digits before the decimal point become zeros, and any that are after the decimal point are dropped. This is called rounding up.

Round to the Nearest Hundred: 3250

Identify the hundreds digit: the 2 in 3250

Identify the next smallest place value: the 5 in 3250

Is that digit greater than or equal to five? Yes, so round up.

Increase the hundreds digit by one, so 2 becomes 3. Every digit after becomes a zero.

3250 rounded to the nearest hundred is 3300

Round to the Nearest Ten: 323.5

Identify the tens digit: the 2 in 323.5

Identify the next smallest place value: the 3 in 323.5

Is that digit greater than or equal to five? No, so round down.

The tens digit stays the same at 2. Every digit after becomes a zero. Digits after the decimal point are dropped.

323.5 rounded to the nearest ten is 320

Round to the Nearest Ten: 499

Identify the tens digit: the first 9 in 499

Identify the next smallest place value: the second 9 in 499

Is that digit greater than or equal to five? Yes, so round up.

The tens digit increases by one. Since 9+1=10, you need to carry the 1 and add it to the digit in the hundreds place. Each digit after becomes a zero.

499 rounded to the nearest ten is 500

Round to the Nearest Tenth: 0.74

Identify the tenths digit: 7 in 0.74

Identify the next smallest place value: the 4 in 0.74

Is that digit greater than or equal to five? No, so round down.

The tenths digit stays the same at 7. Since the remaining digits are after the decimal point you just drop them.

0.74 rounded to the nearest tenth is 0.7

Round to the Nearest Hundredth: 3.141

Identify the hundredths digit: the 4 in 3.141

Identify the next smallest place value: the second 1 in 3.141

Is that digit greater than or equal to five? No, so round down.

The hundredths digit stays the same at 4. Drop the digits to the right of 4.

3.141 rounded to the nearest hundredth is 3.14

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Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

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A. ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and  \sqrt{8^{2} - x^{2} }or  \sqrt{64 - x^{2} }

thus tan(arcsin(x/8))  = \frac{x}{\sqrt{64 - x^{2} }}  =

Therefore  ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B

Here we have cos = adjacent/hypotenuse where adjacent side is \sqrt{64 - x^{2} } and hypothenuse = 8 we have \sqrt{64 - x^{2} }/8

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

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