Want this question answered?
You can't, unless it's an initial value problem. If f(x) is an antiderivative to g(x), then so is f(x) + c, for any c at all.
2x - 3y = 7 -37 + y = 7 The second equation simplifies to -2y = 7 so that y = -7/2 Substituting this value for y in the first equation gives 2x - 3*(-7/2) = 7 2x + 21/2 = 7 or 4x + 21 = 14 so that 4x = -7 and so x = -7/4 So the solution to the given question is (x, y) = (-7/4, -7/2)
It is not to solve so much as to see the number of solutions and whether there is a real solution to the equation. b2 - 4(a)(c) A positive answer = two real solutions. A negative answer = no real solution ( complex solution i ) If zero as the answer there is one real solution.
5x = 15 so x = 15/5 = 3
3x + 5y = -1 2x - 5y = 16 Add the two equations: 5x = 15 or x = 3 Substitute for x in the first equation: 3*3 + 5y = -1 15 + 5y = -1 5y = -10 so y = -10/5 = -2 The solution is (x, y) = (3, -2)
8
Graphs can be used in the following way to estimate the solution of a system of liner equations. After you graph however many equations you have, the point of intersection will be your solution. However, reading the exact solution on a graph may be tricky, so that's why other methods (substitution and elimination) are preferred.
I think you mean the pH value, it so, a basic solution has any value exceeding 7.
It is definatly possible to have a no solution answer in algebra. The most common time a no solution answer will occur is in an absolute value problem. Absolute value problems cannot equal zero so if you have the problem: the absolute value of 2m - 3 = -14 the answer would be so solution because it cannot equal a negative number
Nothing that follows can be in the solution set of 1 so all of it cannot.
7
Designing a system with requirements that maximize performance while minimizing cost and risk would demonstrate the use of the CAIV (Cost As an Independent Variable) philosophy.
The data of the problem lead to the following system of two equations: x + 10y + 50z = 500 x + y + z = 100 We subtract the second equation from the first equation: 9y + 49z = 400 For z = 1, y = 39, x = 60 For z = 2 the system has no solution For z = 3 the system has no solution For z = 4 the system has no solution For z = 5 the system has no solution For z = 6 the system has no solution For z = 7 the system has no solution For z = 8 the system has no solution For z = 9 and higher, y cannot be a positive integer, so we can ignore these. So the only solution would be 1 half dollar, 39 dimes and 60 pennies. That's where I end up after getting 770 Math and 650 Verbal on the SAT and 800 on the Math II achievement test (previous names of the SAT tests). *Sob...*
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
The solution is an acid. If it is in water (supposedly meant by questioneer), the pH value is below 7.0, so it is an acid solution: more H+ than OH-.
That will depend on the plus or minus value of 20 or perhaps there is an equality sign missing and so therefore it has no solution.
5x2 - 2y2 = -20 7x2 - y2 = 152 Eq1 - 2*Eq2: -9x2 = -324 so that x2 = 36 Substituting the value of x2 in Eq1: 2y2 = 200 so that y2 = 100 The four solutions are (-6, -10), (-6, 10), (6, -10) and (6,10)