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ohaa [14]
3 years ago
9

Round to the nearest ten thousand 905154

Mathematics
1 answer:
kolbaska11 [484]3 years ago
5 0
You want to round 905,154 to the nearest ten-thousands place. The ten-thousands place in your number is shown by the bold underlined digit here:

9<em><u>0</u></em>5,154

    To round 905,154 to the nearest ten-thousands place...

    The digit in the ten-thousands place in your number is the 0. To begin the rounding, look at the digit one place to the right of the 0, or the 5, which is in the thousands place.

    Since the 5 is greater than or equal to 5, we'll round our number up by

        Adding 1 to the 0 in the ten-thousands place, making it a 1.

    and by changing all digits to the right of this new 1 into zeros.

The result is: 910,000.

So, 905,154 rounded to the ten-thousands place is 910,000.

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Mariulka [41]

nearest cent so so 32.86 but since its money and 32.86 times 7 is 230.02 it should be rounded down so the answer will be 32.85

5 0
3 years ago
How to factor the gcf out of a polynomial
Ann [662]

Step-by-step explanation:

15x^6+6x^4+9x^3\\\\\text{Find GCF of 15, 6 and 9:}\\\\15=\boxed3\cdot5\\6=\boxed3\cdot2\\9=\boxed3\cdot3\\\\GCF(15,\ 6,\ 9)=\boxed3\\\\\text{Find GCF of}\ x^6,\ x^4\ \text{and}\ x^3:\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\x^6=x^{3+3}=\boxed{x^3}\cdot x^3\\\\x^4=x^{3+1}=x^3\cdot x^1=\boxed{x^3}\cdot x\\\\x^3=\boxed{x^3}\cdot1\\\\GCF(x^6,\ x^4,\ x^3)=\boxed{x^3}\\\\\text{Therefore}\\\\GCF(15x^6+6x^4+9x^3)=\boxed{3}\cdot\boxed{x^3}=3x^3\\\\15x^6+6x^4+9x^3=3x^3(5x^3+2x+3)

8 0
3 years ago
Find the 62nd term of the following sequence:<br> (-21, – 27, - 33, - 39,...)
iris [78.8K]

Answer:

-387

Step-by-step explanation:

First term (T1) = -21 = a

second term (t2) = - 27

common difference (d) = t2 - t1 = - 27 + 21 = - 6

Now

Tn = a + ( n - 1) d

T62 = - 21 + ( 62 - 1 ) - 6

= - 21 -366

= -387

8 0
3 years ago
Function P is defined as P (x) = x ^2 . Explain why P (2) + P (3) isn't equal to P (5) .
diamong [38]

Answer:

P(x) = x + 2

Piden: A = P(P(P(P(3))))

P(3); x = 3

P(3) = 3 + 2 = 5

A = P(P(P(P(3))))

A = P(P(P(5)))

P(5); x = 5

P(5) = 5 + 2 = 7

A = P(P(P(5)))

A = P(P(7))

P(7); x = 7

P(7) = 7 + 2 = 9

A = P(P(7))

A = P(9)

P(9); x = 9

P(9) = 9 + 2 = 11

→ A = P(P(P(P(3)))) = 11

Step-by-step explanation:

listo e

4 0
3 years ago
102 pencils need to be put into packets of 12. How many full packs can be made?
Sunny_sXe [5.5K]

Answer:

Full packs = 8

Left over = 6 pencils

Step-by-step explanation:

102 divided by 12 = 8

with a remainder of 6

5 0
2 years ago
Read 2 more answers
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