-4q^4 +0q^3+2q^2-0q-1 -(0q^4+q^3-5q^2+7q-9)=

-4q^4 +0q^3+2q^2-0q-1 -q^3+5q^2-7q+9=

-**4q^4-q^3+7q^2-7q+8**

Posted on Nov 25, 2013

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Posted on Jan 02, 2017

- Make sure you're working with two fractions.
- Multiply numerator x numerator, then multiply denominator x denominator.

- Make sure you're working with two fractions
- Flip the second fraction upside down.
- Change the division sign into a multiplication sign.
- Multiply top x top and bottom x bottom.

- Convert mixed numbers into improper fractions.
- Take the whole (non-fraction) number and multiply it by the denominator.
- Add that answer to the numerator.
- Put that amount over the original denominator and you will have an improper fraction.

- Find the lowest common denominator (bottom number). For both adding and subtracting fractions, you'll start with the same process.
- Multiply fractions to match the lowest common denominator.
- Add or subtract the two numerators (top number) but NOT the denominators.

Mar 07, 2017 | Homework

Decimal is base 10 and binary is base 2.

The columns from right to left in base 10 are:

10^10 10^9 10^8 10^7 10^6 10^5 10^4 10^3 10^2 10^1 10^0

Similarly, the columns from right to left in base 2 are:

2^10 2^9 2^8 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0

This site does a good job of explaining the conversion from decimal to binary.

http://www.math.grin.edu/~rebelsky/Courses/152/97F/Readings/student-binary

For example, take the number 157. The biggest column that goes into 157 is 128, which is the 2^7th column, so we put 1 in that column. Subtracting 128 from157, we get 29. We start all over again. The biggest binary number going into 29 is 16, which is the 2^4th column. When you subtract 16 from 29, you get 13. We start all over again. The biggest binary number going into 13 is 8. Subtracting 8 from 13, we get 5. Biggest number going in is 4, 2^2th column. 1 left over to go into the 2^0 column.

Pulling it all together, we get:

10011101.

Checking, 128 + 16 + 8 + 4 + 1 = 157.

Good luck.

Let me know if you have any questions.

Paul

The columns from right to left in base 10 are:

10^10 10^9 10^8 10^7 10^6 10^5 10^4 10^3 10^2 10^1 10^0

Similarly, the columns from right to left in base 2 are:

2^10 2^9 2^8 2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0

This site does a good job of explaining the conversion from decimal to binary.

http://www.math.grin.edu/~rebelsky/Courses/152/97F/Readings/student-binary

For example, take the number 157. The biggest column that goes into 157 is 128, which is the 2^7th column, so we put 1 in that column. Subtracting 128 from157, we get 29. We start all over again. The biggest binary number going into 29 is 16, which is the 2^4th column. When you subtract 16 from 29, you get 13. We start all over again. The biggest binary number going into 13 is 8. Subtracting 8 from 13, we get 5. Biggest number going in is 4, 2^2th column. 1 left over to go into the 2^0 column.

Pulling it all together, we get:

10011101.

Checking, 128 + 16 + 8 + 4 + 1 = 157.

Good luck.

Let me know if you have any questions.

Paul

May 14, 2016 | Office Equipment & Supplies

get the X's on the same side of the equation by subtracting 3x from both sides, then get the constants on the same side by subtracting 5 from each side. You'll simplify it to 6x-3x=17-5, resulting in 3x=12 or x=4...

Nov 14, 2014 | Computers & Internet

SUM means adding not subtracting. so if you were wanting to add the total (the sum) of the amounts of the hundreds and the tens would be 170 plus 130 which would be 300 but the remainder from subtracting 130 from 170 would be 40. That would leave the single units 5 minus 4 of which the remainder is 1. So, the remainder from subtracting 134 from 175 is the reminder of subtracting the one hundred from the one hundred which is zero added to the remainder of subtracting the tens (70 minus 30) which is the remainder of 40 plus the remainder of subtracting the units of 5 minus 4 which is 1. So you need to add the 0 and the 40 and 1 which is the remaining sum of 41.

Sep 19, 2014 | Office Equipment & Supplies

Hi there,

Remember that the SHARP calculator does not solve algebra.

But some tips for adding and subtracting monomials include:

Group all the terms that are like terms (they have the same variables and exponents. Make sure that you move them with the sign in front of them (dont lose your minuses).

Add or subtract just the numbers in front of the terms and leave the variables exactly as they are when you write down the answer.

Remember that the SHARP calculator does not solve algebra.

But some tips for adding and subtracting monomials include:

Group all the terms that are like terms (they have the same variables and exponents. Make sure that you move them with the sign in front of them (dont lose your minuses).

Add or subtract just the numbers in front of the terms and leave the variables exactly as they are when you write down the answer.

Feb 20, 2014 | Sharp el-531x scientific calculator

log3(u^2/v)

You can move the top as an exponent because of the rules of logs. Then subtracting two same base logs can be combined because another rule. If they are subtracting you divide them and if they adding you multiply them.

You can move the top as an exponent because of the rules of logs. Then subtracting two same base logs can be combined because another rule. If they are subtracting you divide them and if they adding you multiply them.

Apr 23, 2010 | MPS Multimedia Speedstudy Pre Calculus...

6x+6=4x+12

Since 4x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 4x from both sides.

6x+6-4x=12

Since 6x and -4x are like terms, add -4x to 6x to get 2x.

2x+6=12

Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.

2x=-6+12

Add 12 to -6 to get 6.

2x=6

Divide each term in the equation by 2.

(2x)/(2)=(6)/(2)

Simplify the left-hand side of the equation by canceling the common factors.

x=(6)/(2)

Simplify the right-hand side of the equation by simplifying each term.

x=3

Good Luck

Since 4x contains the variable to solve for, move it to the left-hand side of the equation by subtracting 4x from both sides.

6x+6-4x=12

Since 6x and -4x are like terms, add -4x to 6x to get 2x.

2x+6=12

Since 6 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 6 from both sides.

2x=-6+12

Add 12 to -6 to get 6.

2x=6

Divide each term in the equation by 2.

(2x)/(2)=(6)/(2)

Simplify the left-hand side of the equation by canceling the common factors.

x=(6)/(2)

Simplify the right-hand side of the equation by simplifying each term.

x=3

Good Luck

Sep 10, 2009 | Audio Players & Recorders

Hello desireejane,

One method is to do the following

public static long octalToDecimal(String octal) throws NumberFormatException {

// Initialize result to 0

long res = 0;

// Do not continue on an empty string

if (octal.isEmpty()) {

throw new NumberFormatException("Empty string is not an octal number");

}

// Consider each digit in the string

for (int i = 0; i < octal.length(); i++) {

// Get the nth char from the right (first = 0)

char n = octal.charAt(octal.length() - (i+1));

int f = (int) n - 48;

// Check if it's a valid bit

if (f < 0 || f > 7) {

// And if not, die horribly

throw new NumberFormatException("Not an octal number");

} else {

// Only add the value if it's a 1

res += f*Math.round(Math.pow(2.0, (3*i)));

}

}

return res;

}

One method is to do the following

- Convert the octal, hexadecimal or binary to decimal.
- Add or Subtract the decimal normally
- Convert the result back to octal, hexadecimal or binary.

**Convert the octal to decimal:**

public static long octalToDecimal(String octal) throws NumberFormatException {

// Initialize result to 0

long res = 0;

// Do not continue on an empty string

if (octal.isEmpty()) {

throw new NumberFormatException("Empty string is not an octal number");

}

// Consider each digit in the string

for (int i = 0; i < octal.length(); i++) {

// Get the nth char from the right (first = 0)

char n = octal.charAt(octal.length() - (i+1));

int f = (int) n - 48;

// Check if it's a valid bit

if (f < 0 || f > 7) {

// And if not, die horribly

throw new NumberFormatException("Not an octal number");

} else {

// Only add the value if it's a 1

res += f*Math.round(Math.pow(2.0, (3*i)));

}

}

return res;

}

Mar 24, 2009 | Sun Java Programming Language (cdj-275)

can you example more pls

Oct 01, 2008 | Cell Phones

Because this is an adding machine you would use the total button (*) not the = for ading and subtracting the = is only used in multiplcation and division

Apr 12, 2008 | Canon P23-DH Calculator

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