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How do you make a 5 X solution from 10 X stock?
To make a 5X solution from a 10X stock, you can dilute the 10X stock solution by adding an equal volume of diluent (such as water or buffer) to the original solution. For example, if you have 1 mL of the 10X stock solution, you would add 1 mL of diluent to make a 5X solution.
What does it mean to evaluate an equation?
It means that you find the specific value of the equation substituting your specific variables. for example: equation: z=2y+x-3 the evaluatio of the equation for x=1 and y=3 will be: z=2(3)+1-3=4
Is 1-3 a solution to 3y 2x 1?
I assume you mean 3y+2x+1. So sub in the points and see. The 1 is x and the -3 is y. 3(-3) + 2(1) + 1. Does this equal 0? -9+2+1 equals -6, so no, it is not a solution. For example, if your point is (1,-1), this would be a solution.
What is the molarity of an hcl solution if 7ml hcl solution is titrated with 27.6ml of 0.170m of naoh solution?
The balanced chemical equation for the reaction is: HCl + NaOH -> NaCl + H2O. From the equation, it is a 1:1 mole ratio reaction. Therefore, the moles of HCl can be calculated from the volume and concentration of NaOH used in the titration. Then, use the moles of HCl and the volume of HCl solution used to calculate the molarity of the HCl solution.
How do you prepare 1 M solution from 6 M solution?
It's simple to dilute a solution, just add more water. If you have one liter of 6 molar solution, and add another 5 liters of water, it becomes a 1 molar solution. You can also use M1V1 = M2V2 formula for dilution. For example, if you want to prepare 1 M of solution in 1 litre of water, then how much volume you need from the initial 6 M solution? Simply use the equation, (6 M) x (V1) = (1 M) x (1 litre) ---> V1 = (1 M) x (1 litre)/(6 M) = 0.167 L or 167 ml
What is a solution to the equation 5x 2y-1?
Without an equality sign it cannot be classed as an equation therefore it follows that there is no solution.
What is equation of 3x plus 2y equals 8 and 5x plus 2y equals 12?
Both of them are equations!The solution is (x, y) = (2, 1).
What is the solution of 2y-x equals 3 and y equals -5x plus 7?
2y - x = 3 y = -5x + 7 Rewrite the second y + 5x = 7 Mutiply it by 2 2y + 10x = 14 Now subtract the first equation from this: 2y + 10x = 14 2y - x = 3 11x = 11 Divide both sides by 11: x = 1 Substitute for x in the first equation: 2y - 1 = 3 Add 1 to both sides: 2y = 4 Divide both sides by 2: y = 2 Answer: (x, y) = (1, 2)
5x plus 3y equals 17 5x - 2y equals -3?
Multiply the bottom equation by -1
What is the solution to 2y equals 5x - 1 x equals y plus 2?
If those are two different equations, such that 2y=5x-1 and x=y+2, then you subsitute y+2 in for x in the first equation, and when you solve the whole thing, you get that x=-1 and y=-3, or the point (-1,-3)
How do you solve x y0 5x 2y -3?
To solve the equation ( xy^0 + 5x + 2y - 3 = 0 ), note that ( y^0 = 1 ), simplifying the equation to ( x + 5x + 2y - 3 = 0 ) or ( 6x + 2y - 3 = 0 ). Rearranging gives ( 2y = 3 - 6x ), leading to ( y = \frac{3 - 6x}{2} ). This represents a linear relationship between ( x ) and ( y ).
Solve the following system of equations using substitutions 2x plus 3y equals -1 and 5x-2y equals -12?
First get y in terms of x: 5x-2y=-12 ----- -2y=-12-5x ------ y=6+(5/2)x Sub into other equation: 2x+3y=-1 ---- 2x+3(6+(5/2)x)=-1 Solving for x gets: x= -2 Sub into other equation: 5x-2y= -12 --- 5(-2)-2y= -12 --- y=1 So x= -2, and y= 1
Solve this equation using substitution 5x plus y equals 1 and 3x plus 2y plus 2?
It is not possible to do so because the question contains one equation (5x + y = 1) and one expression (3x + 2y + 2). An expression cannot be solved.
Isha sharma writes three liner equations in two variables as under 3x plus 2y equals 5 5x-7y equals -2 6x plus 4y equals 10which pairs has no solution?
There is no such pair. The solution to equation 1 and equation 2 is x = 1, y = 1. The solution to equation 2 and equation 3 is x = 1, y = 1. And the solution to equation 1 and equation 3 is any point on the line 3x + 2y = 5 - an infinite number of solutions. The fact that the determinant for equations 1 and 3 is zero (or that they are not independent) does not mean that there is no solution. It means that there is no UNIQUE solution. In this particular case, the two equations are equivalent and so have an infinite number of solutions.
5x plus y equals 5 3x plus 2y equals 3?
Double first equation: 10x + 2y = 10Subtract second equation: 7x = 7x = 1, y = 0
3x plus 2y equals 20 and 5x-2y equals 1?
3x+2y=20 5x-2y=1 (3x+2y) + (5x-2y)= 20+1 3x+5x= 21....... 8x=21.........x=2 5/8 Subsitute x=3(2 5/8)+ 2y=20........... 7 7/8 +2y=20.........2y= 20- 7 7/8....... y=6 1/16
What Is 2y plus 4a equals?
x-2y+bz=3 ax+2z=2 5x+2y=1 find values of a and b for which the system of linear equation has infinitely many solutions
