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Blababa [14]
3 years ago
11

How do I subtract decimals from whole numbers?

Mathematics
1 answer:
Gekata [30.6K]3 years ago
7 0

Answer:

Just use long subtraction by expanding the decimal places of the whole number. This is done by adding a point, and enough zeros to it to match the number of decimal digits in the other number (digits after the decimal point).

12345678

i.e: 5 - 2.48374827, 2.48374827 has 8 decimal digits, so add 8 zeros after the point.

=

1 1 1 1 1 1 1

5.00000000

-

2.48374827

_______________

2.51625173

7 + 3 = <u>1</u>0, 7 + 2 + <u>1</u> = <u>1</u>0, 8 + 1 + <u>1</u>= <u>1</u>0, 5 + 4 + <u>1</u> = <u>1</u>0, 2 + 7 + <u>1</u> = <u>1</u>0, 3 + 6 + <u>1</u> = <u>1</u>0, 1 + 8 + <u>1</u> = <u>1</u>0, 5 + 4 + <u>1</u> = <u>1</u>0, 2 + 2 + <u>1</u> = <u>5</u><u> </u><u>:</u><u> </u>5.00000000

This is basically borrowing a group of 10s which are the same as 1s in the next decimal place up.

For each digit except the first to the right, let 10 subtract that number from it and minus 1 since the 1 is carried over.

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What is the factored form of y=x^3+3x^2-x-3 and what the zeros
PSYCHO15rus [73]
   y = x³ + 3x² - x - 3
   0 = x³ + 3x² - x - 3
   0 = x²(x) + x²(3) - 1(x) - 1(3)
   0 = x²(x + 3) - 1(x + 3)
   0 = (x² - 1)(x + 3)
   0 = (x² + x - x - 1)(x + 3)
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   0 = x - 1    or    0 = x + 1    or    0 = x + 3
+ 1      + 1         - 1        - 1         - 3        - 3
   1 = x      or      -1 = x       or      -3 = x

Solution Set: {-3, -1, 1}
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-54 + 108 =<br> Thank you!
julsineya [31]

Answer: 54 and welcome

Step-by-step explanation:

3 0
3 years ago
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Error analysis
Bas_tet [7]
2(2x-1) + 2(3x)=4x-2+6x = 10x - 2 & not <span>=10x-1</span>
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3 years ago
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I dont understand this
Westkost [7]

Answer:

\dfrac{1}{2x(x-1)}

Step-by-step explanation:

Given

\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)

Consider the numerator:

x^2+2x+1=(x+1)^2

Consider the denominator:

x^2-1=(x-1)(x+1)

Hence, the fraction becomes

\dfrac{(x+1)^2}{(x-1)(x+1)}=\dfrac{x+1}{x-1}

Consider the expression in brackets:

2x^2+2x=2x(x+1)

Divide:

\dfrac{x^2+2x+1}{x^2-1}\div (2x^2+2x)=\dfrac{x+1}{x-1}\div 2x(x+1)=\dfrac{x+1}{x-1}\times \dfrac{1}{2x(x+1)}=\dfrac{1}{2x(x-1)}

4 0
3 years ago
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