Round 452,906 to the nearest hundred thousand, ten thousand, thousand, hundred, ten

Mathematics

2 answers:

OLEGan [10]3 years ago

8 0

452,906

Nearest hundred thousand : 500,000
Nearest ten thousand : 450,000
Nearest thousand : 453,000
Nearest hundred: 452,900
Nearest ten: 452,910

If the place value below the place you’re rounding to is above 5 round up if below round down.

meriva3 years ago

4 0

Answer:
500,000
450,000
453,000
452,900
452,910

Explanation:
Look to the right one digit of the place they are asking to round up to. If it’s either greater than or equal to 5, then you round up the place they want. If it’s not, then it stays the same. Then all the number to right of it become 0s.

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a. On a hiking trip Bob and his two friends hiked 15 miles the first day, 10 miles on the second and 7 miles on the third day. I

Len [333]

Answer: A. 14 miles each day.

Step-by-step explanation:

Day 1 : 15 miles

Day 2 : 10 miles

Day 3 : 7 miles

15 + 10 + 7 = 32 miles

60 - 32 = 28 miles

2 days left

28/2 = 14

14 is less than 18

7 0

3 years ago

An electronics store makes a profit of $72.00 for every television sold and $90.00 for every computer sold. The manager’s target

viktelen [127]

Answer:

<h2>Option A, C and D is the three possible solution.</h2>

Step-by-step explanation:

Each Television earns $72 and each Computer earns $90.

The manager’s target is to make at least $360.00 a day.

It is <u>not clarified the exact number of Televisions and Computers</u>.

Hence, as per the given condition, $72x + 90y \geq 360\\8x + 10y \geq 40\\4x + 5y \geq 20$ .

Now we just need to check the given options, whether they satisfy the above equation, or not.

Except the option B, each inequality indicates a value greater than 360.

Hence, option B can not be the possible solution.

4 0

3 years ago

765ab4 is divisible by 36. Compute all ordered pairs (a, b) for which this is possible

tia_tia [17]

Answer:

All the ordered pairs (a,b) are

(0,5), (2,3), (4,1), (6,8), and (8,6).

Step-by-step explanation:

765ab4 to be divisible by 36, it must be divisible by both 9 and 4.

For any number to be divisible by 4, the last two digits of the number must be divisible by 4.

Here, the last two digits are b4, so as the number b4 is divisible by 4, b can be 0,2,4,6, and 8.

For any number to be divisible by 9, the sum of all the digit of the number must be divisible by 9.

Here, the sum of all the digits, S= 7+6+5+a+b+4=22+a+b.

Now, we have b={0,2,4,6, 8}

For b=0,

S=22+a+0=22+a which is a multiple of 9.