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3 years ago

How do you write 7 1 \12 as a whole number

Mathematics

1 answer:

Alex777 [14]3 years ago

3 0

Answer: 7

Step-by-step explanation:

Write as a decimal

7 1/2 would be written as ~7.0833

Rounding a decimal to the nearest whole number

If you have a decimal that is greater than or equal to 5 in the tenths place then you round up. Ex: 5.8 would be rounded to 6, 1.5 would be rounded to 2.

If you have a decimal that is less than or equal to 4 in the tenths place then you round down. Ex: 5.1 would be rounded 5, 1.2 would be rounded to 1.

Since 0 is in the tenth’s place we round down to 7.

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Margarita [4]

Answer:

It comes down to what the equals sign means. It means, that the two things mentioned on either side of the equation are the same thing. If you do something to the left side, you have presumably changed it.

Step-by-step explanation:

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3 years ago

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7,000,000 + 600,000 + 70,000 + 2,000 + 500 + 30 + 1 in standard form?

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4 years ago

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Which decimal number is equivalent to 6/25?<br> A.0.6<br> B.0.12<br> C.0.06<br> D.0.24

blondinia [14]

D if you take 6/25 = 24/100 then change to a decimal and get 0.24

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3 years ago

Find the common fraction equivalent to 0.12

jonny [76]

The answer is: " $\frac{4}{33}$ "  .
______________________________
Given: 0.1212121212..... repeating ; write that value as a fraction;
______________________________________________________
In other words; we are given: "0.1212121212..... repeating infinitely" ;

→ that is to say; "0.12 ... ; {the "12" decimal portion repeats infinitely} ;
_______________________________________________________
→ We write this value, as a fraction, as: "12/99" ;
__________________________________________
→Explanation:
__________________________________________

Note: "0.99999999...... repeating infinitely; = "1" .
_________________________________________
→Since:

Let us say that we have:

"x = 0.999999 ; repeating infinitely;

In order words, let us say we have: "x = 0.9.... ; the "9" decimal repeats infinitely;
_____________________________________________________
Then "10x" ; (that is: "10" multlipled by "x"; or "10*x" or "10x" ); is equal to:

"10" * (0.999999.....) = 9.99999999...... (the "9" decimal repeats infinitely);

in other words: 10x = 9.99999999....

Divide each side by "10" ;

to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;

But if you have: "10x = 10" ; divide each side of the equation by "10" ;
you get: "x = 1" .
____________________
Also, if "x = 0.9999...(repeating infinitely);

then: 10x = 9.99999.
_______________________________________________
10x = 9.999999999999999......
−   x = 0.999999999999999.......
_____________________________________
9x = 9.00000000000000000000.....

→ 9x = 9 ;

Divide each side of the equation; to get;

9x /9 = 9/ 9 ; to get: x = 1 ; and we have: x = 0.9999.... ; so
x= 0.99999.... = 1 ;
__________________________________________________
So, if the numbers "12" is repeating, we divie "12" by "99" ;
that is; we divide "12" by "two 9's" ; since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
________________________________________
So; 0.12121212.....(the "12" is the decimal that repeats infinitely);

= 12/99 ; which can be simplified;

Divide each side (both the numerator AND the denominator); by "3" ;
_________________________________________
" 12/99 " = "(12÷3) / (99÷3) = 4/33 " .
_________________________________________
The answer is: $\frac{4}{33}$   .
_________________________________________

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3 years ago

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PLEASE HELP WITH GRADE 11 MATH. Make sure you show the formula. Substitute values and show all mathematicall operations! show yo

Degger [83]

3x+y

=−3

=−y+3

The second equation is solved for xxx, so we can substitute the expression -y+3−y+3minus, y, plus, 3 in for xxx in the first equation:

\begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}

3x+y

3(−y+3)+y

−3y+9+y

−2y

=−3

=−12

Plugging this value back into one of our original equations, say x = -y +3x=−y+3x, equals, minus, y, plus, 3, we solve for the other variable:

\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}

=−y+3

=−(6)+3

=−3

The solution to the system of equations is x=-3x=−3x, equals, minus, 3, y=6y=6y, equals, 6.

We can check our work by plugging these numbers back into the original equations. Let's try 3x+y = -33x+y=−33, x, plus, y, equals, minus, 3.

\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}

3x+y

3(−3)+6

−9+6

−3

=−3

−3

=−3

Yes, our solution checks out.

Example 2

We're asked to solve this system of equations:

\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}

7x+10y

−2x+y

=36

In order to use the substitution method, we'll need to solve for either xxx or yyy in one of the equations. Let's solve for yyy in the second equation:

\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}

−2x+y

=2x+9

Now we can substitute the expression 2x+92x+92, x, plus, 9 in for yyy in the first equation of our system:

\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}

7x+10y

7x+10(2x+9)

7x+20x+90

27x+90

3x+10

=36

=−6

=−2

Plugging this value back into one of our original equations, say y=2x+9y=2x+9y, equals, 2, x, plus, 9, we solve for the other variable:

\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}

=2x+9

=2(−2)+9

=−4+9

The solution to the system of equations is x=-2x=−2x, equals, minus, 2, y=5y=5y, equals, 5.

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2 years ago