What is 694 divided by 41 and what is the remainder please help thank you
2 answers:
We can use long division in finding the remainder.
Place the divisor which is 41, outside the long division sign, and 694 under the long division sign as shown in the diagram.
41 goes into 694 16 times.
Write the 16 on top of the long division sign.
Multiply the 16 by 41, to obtain 656
Write the 656 below the 694 as in the diagram.
Subtract 656 from 694 to obtain 38.
Since 41 is more than 38 , we cannot proceed.
Hence,
Therefore the remainder is 38
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<span>1.1 What are the different ways you can solve a system of linear
equations in two variables?
</span><span>A linear equation can be written in many ways, so the way we can use to solve a system of
linear equations depends on the form the system is written. There are
several methods to do this, the more common are: Method of Equalization,
Substitution Method, Elimination Method.
<span>1.2. Method of Equalization
</span>a. Write the two equations in the
style [variable = other terms] either variable x or y.<span>
</span>b. Equalize the two equations
c. Solve for the other variable and then for the
first variable.
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1.3. Substitution Method
a. Write one of the equations (the one that looks the simplest equation) in
the style [variable = other terms] either variable x or y.
b. Substitute that variable in the other equation and solve using the
usual algebra methods.
c. Solve the other equation for the other variable.<span>
1.4. Elimination Method
Eliminate</span> means to remove, so this method works by removing
variables until there is just one left. The process is:
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</span>a. Multiply an equation by a constant "a" such that there is a term in one equation like [a*variable] and there is a term in the other equation like [-a*variable]
b. Add (or subtract) an equation on to another equation so the aim is to
eliminate the term (a*variable)
c. Solving for one variable and the for the other.
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2.1 Benefits of Method of Equalization<span>
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It's useful for equations that are in the form [variable = other terms] (both equations). In this way, it is fast to solve the system of linear equation by using this method. The limitations come up as neither of the equations are written in that way, so it would be necessary to rewrite the two equations to achieve our goal
2.2. Benefits and limitations of Substitution Method
This method is useful for equations that at least one of them is in the form [variable = other terms]. So unlike the previous method, you only need one equation to be expressed in this way. Hence, it is fast to solve the system of linear equation by using this method. The limitations are the same that happens with the previous method, if neither of the equations is written in that way, we would need to rewrite one equation to achieve our goal.
2.3. Benefits and limitations of Elimination Method
It's useful for equations in which the term [a*variable] appears in one equation and the term [-a*variable] appears in the other one, so adding (or subtracting).
3.1. What are the different types of solutions
When the number of equations is the same as the number of variables there is likely to be a solution. Not guaranteed, but likely.
One solution: It's also called a consistent system. It happens as each equation gives new information, also called Linear Independence.
An infinite number of solutions: It's also called a consistent system, but happens when the two equations are really the same, also called Linear Dependence.
No solutions: It happens when they are actually parallel lines.
3.2. Graph of one solution system
See figure 1, so there must be two straight lines with different slopes.
3.3. Graph of infinite number of solutions system
See figure 2. So there must be two straight lines that are really the same.
3.4. Graph of no solution system
See figure 3. So there must be two straight lines that are parallel.
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If you rent a car from the first car agency your cost for the rental will be:
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Given that the problem says you want to rent the car just for one day, then D = 1, therefore:
First agency:
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There is a representation of this problem in figure 4. Up until you drive 200
miles you would save money by going with the second company.</span>
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solved by a system of equations in two variables </span>
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