It is: (2x-5)(2x-5) = 2.5 square cm when solved as a quadratic equation
Equation: 10x^2 -29x +10 = 0 When factored: (2x-5)(5x-2) = 0 Its solutions: x = 5/2 or x = 2/5
Known equation: 3x+5y = 6 or y = -3/5x +6/5 Slope of equation: -3/5 Slope of parallel equation: -3/5 Parallel equation: y-1 = -3/5(x-3) => 5y = -3x+14 Parallel equation in its general form: 3x+5y-14 = 0
L= Length W= Width P= Perimeter Equation 1: L= 2W-5 Equation 2: 2L+2W=P=80 Then, From Equation 2, Solve the second equation for 2W. 2L + 2W= 80 2W = 80 - 2L From Equation 1, Substitute 80-2L for 2W in the first equation. This gives the equation one variable, which earlier algebra work.L=(80-2L)-5 L=80-2L-5 2L + L= 80 - 5 3L= 75 L=25 Now, substitute 25 for L in either equation and solve for w. From Equation 1 25 = 2W - 5 5 +25 = 2W30 = 2W 30 / 2 =W 15=W The solution is Lengh = 25 Width = 15 L= Length W= Width P= Perimeter Equation 1: L= 2W-5 Equation 2: 2L+2W=P=80 Then, From Equation 2, Solve the second equation for 2W. 2L + 2W= 80 2W = 80 - 2L From Equation 1, Substitute 80-2L for 2W in the first equation. This gives the equation one variable, which earlier algebra work.L=(80-2L)-5 L=80-2L-5 2L + L= 80 - 5 3L= 75 L=25 Now, substitute 25 for L in either equation and solve for w. From Equation 1 25 = 2W - 5 5 +25 = 2W30 = 2W 30 / 2 =W 15=W The solution is Lengh = 25 Width = 15
Since we know the slope, m = 5/3, and the y-intercept 1/2, we arw able to write the equation of the line in the slope-intercept form, y = mx + b, so we have y = (5/3)x + 1/2.The standard form of the equation of the line is Ax + By = C.y = (5/3)x + 1/2y - y - 1/2 = (5/3)x - y + 1/2 - 1/2-1/2 = (5/3)x - y or(5/3)x - y = -1/2Thus, the standard form, Ax + By = C, of the equation of the line is (5/3)x - y = -1/2.
x2 - 12x + 35
The discriminant is 36 which means the quadratic equation has two solutions which are 5 and -1
(x + 5) (x + 1) = 0x2 + 6x + 5 = 0
x2 - 8x + 15
The given expression is a quadratic equation. To find its solutions, we can either factor the equation or use the quadratic formula. However, without an equation to solve or any context, it is not possible to provide a numeric answer.
This is a quadratic equation which will have two solutions: X2 = 4x+5 Rearrange the equation: x2-4x-5 = 0 Factor the equation: (x+1)(x-5) = 0 So the solutions are: x = -1 or x = 5
There are no real solutions because the discriminant of the quadratic equation is less than zero.
The number of solutions an equation has depends on the nature of the equation. A linear equation typically has one solution, a quadratic equation can have two solutions, and a cubic equation can have three solutions. However, equations can also have no solution or an infinite number of solutions depending on the specific values and relationships within the equation. It is important to analyze the equation and its characteristics to determine the number of solutions accurately.
x2 + 4x = 41
It is a quadratic equation in the form of y2-4y-5 = 0 and will have two solutions: When factorised: (y-5)(y+1) = 0 Therefore: y = 5 or y = -1
Translate to what? I assume you need help interpreting it. The quadratic equation is used to solve the quadratic polynomial, ax2 + bx + c = 0, where a, b, and c can be any number. For example, if you need to solve the equation x2 = 5 + 2x, you first convert it into the standard form mentioned above: x2 - 2x - 5 = 0. Now find the coefficients, a, b, and c. In this case, a = 1, b = -2, c = -5. Finally, you replace these coefficients in the quadratic equation. The "plus-minus" sign simply means that the quadratic equation is a shortcut for two equations - one in which you add, the other in which you subtract, the terms at the top. The solutions given by the quadratic equation are values of "x" that satisfy the equation.
2x2-10+7 = 0 Solving the quadratic equation using the quadratic formula will give you two solutions and they are: x = (5 - the square root of 11)/2 or x = (5 + the square root of 11)/2