The equation
7 x + 14 y + 10 z = 98001
has infinite solutions. For any x and y we can have a z which satisfies this equation.
For example, if we choose x=1 and y=1 then z=9798 satisfies.
To have a specific solution there must be three equations for three unknowns (x,y,z), which can be solved simultaneously.
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To solve this problem, we first need to establish the relationship between x, y, and z. From the given information, we have 15x = 20y and 16y = 10z. To find the relationship between x and z, we can substitute y from the first equation into the second equation. By doing this, we get 16(15/20)x = 10z, which simplifies to 12x = 10z. Therefore, 6x would equal 5z.
In algebra it is simply 10z
To make a pound using only 50p, 20p, and 10p coins, we can set up a linear Diophantine equation: 50x + 20y + 10z = 100, where x, y, and z are the number of each type of coin used. We can then use techniques such as generating functions or modular arithmetic to solve this equation and find the number of ways to make a pound. The solution will involve finding the integer solutions to the equation within a certain range, as there are constraints on the number of each type of coin available.
The coins we know are:A penny - 1 centA nickel - 5 centsA dime - 10 centsA quarter - 25 centsLet x denote the number of pennies, y denote the number of nickels, z denote the number of dimes and w denote the number of quarter.We can set up the expression this way:x + 5y + 10z + 25w = 14But since we want to make 14 cents, quarters are not considered to be the part of the combination, so we write:x + 5y + 10z = 14There is no quick way to work out this problem. All you need to do is to work by trials.If we have y = 0 nickels and z = 1 dime, then the remaining choice is 4 pennies to make 14 cents.If we have y = 2 nickels and z = 0 dimes, then the remaining choice is also 4 pennies.If we have y = 1 nickel and z = 0 dimes, then the remaining choice is 9 pennies.If y = z = 0 of these coins - nickels and dimes - , then we can make 14 cents by 14 pennies.So there are 4 ways to form 14 cents.
11z+√(4z)+√(10z)=11z+√(10z)+2*√(z)
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y2 + 10z - 10y - yz = y2 - 10y - yz + 10z = y(y - 10) - z(y - 10) = (y - 10)(y - z)
-6z 2 = 3z+ 4z+28 -6z+2 +3z = 3z+ 4z +28 +4z -6z+4z+2 = 3z+ 28 -10z+2 -3z= 28 - 2 -10z + -3z = 26 -13z/13 = 26/-13 z = -2
Ten z plus one, or ten z plus ten. That depends on where you had parentheses in the original expression: (10 * z) +1 = 10z + 110 * (z + 1) = 10z + 10
To find the LCM, you first need to express the numbers as the product of their prime factors. In this case: 10z = 2x5xz 20x = 2x2x5xz The next step is to identify the HCF. In this case that's 2x5xz = 10z. Then you multiply the numbers together and divide by the HCF: 10z x 20z/10z = 20z Thus the LCM of 10z and 20z is 20z.
Since 20z is a multiple of 10z, it is automatically the LCM of this problem.
To solve this problem, we first need to establish the relationship between x, y, and z. From the given information, we have 15x = 20y and 16y = 10z. To find the relationship between x and z, we can substitute y from the first equation into the second equation. By doing this, we get 16(15/20)x = 10z, which simplifies to 12x = 10z. Therefore, 6x would equal 5z.
4
In algebra it is simply 10z
(x+5)(y-2z)
We can simplify this expression by combining the like terms. Here the likes terms are the z's and the x's.2z + 3z + 5z = 10z. (If we have 2 of something and add three of the same thing and then 5 of the same thing we will end up with 10 of that thing).Likewise we can combine the x's.6x - 2x = 4x. (You could think of this as 6x + - 2x if this helps with the idea of "combining".)Therefore we are able to simplify this expression as:2z + 3z + 5z + 4 + 6x - 2x = 10z + 4 + 4x.