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3 years ago

14

A whole number rounded to the nearest ten thousand is 620,000. What is the least possible value for that number?

1 answer:

Ymorist [56]3 years ago

4 0

Answer:

615,000

Step-by-step explanation:

Given that:

Whole number rounded to the nearest ten thousand is 620,000

Let x = the whole number

HTh__ TTh ____ Th ____ H ____ T ____ U

6_____ 2 ______ 0 ____ 0 ____ 0 ____ 0

The ten thousand digit is 0 ;

To round to the nearest 10 thousand, the thousand digit is rounded up to 1 and added to the ten thousand digit if it up to 5 or above and rounded down to 0 if below 5 with all subsequent digits(hundred, tens and unit) rounded to Zero..

Hence,

The whole number will could be range between :

615,000 - 619,999

Hence, the least possible value for the number is 615,000.

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Solve the following system

scZoUnD [109]

Answer:

{x = -4 , y = 2 , z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = -4 , y = 2 , z = 1

4 0

3 years ago

Show algebriacally that the sum of two consecutive numbers is always odd

egoroff_w [7]

X + (x+1) = 2x+1
Here, 2x+1 is always odd
For example the value of x is 2 then we can say that,
x=2; x+1=3; 2x+1 = 5 <u>[odd number(5)]</u>
= 2 + 3 = 5

<em>Please note that the bolded words are only for your better understanding</em>

HOPE IT HELPS!

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3 years ago

5 ports Use absolute value to express the distance between -12 and -15 on the number line. |-12 - (-15) = -27

zloy xaker [14]

Answer:

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Step-by-step explanation:

Given

-12 and -15

Required

Solve the difference

The question you posted included the solution. However, the solution is incorrect;

The absolute difference is represented as:

$Difference = \vert a - b \vert$

In this case:

$a = -12$

$b = -15$

So, the distance is calculated as:

$Distance = |-12 - (-15)|$

Open bracket

$Distance = |-12 +15|$

$Distance = |3|$

$Distance = 3$

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Answer:

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Step-by-step explanation:

Remark

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The opposite side is not defining the angle. The hypotenuse is the longest side of a right triangle.

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Step-by-step explanation:

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