Yes that about sums it up.
Assuming the equation of the given line is y = 3x + 4, its gradient is 3.Therefore the gradient of the required line is 3. The line passes through (3, 1) and so its equation is y - 1 = 3*(x - 3) = 3x - 9 So y = 3x - 8
The standard equation of a straight line in the slope-intercept form is y = mx + c where m is the slope and c is the y intercept.3x - 6y = 12 can be re-written as 6y = 3x - 12 or y = 1/2x - 2.-2 is therefore the y intercept (this is when x = 0)The x intercept (when y = 0) can be obtained from the equation y = 1/2x - 2 : 0 = 1/2x - 2 : 1/2x = 2 : x = 4.
Definition of slope intercept form:The slope-intercept form is one way to write a linear equation (the equation of a line). The slope-intercept form is written as y = mx+b, where m is the slope and b is the y-intercept (the point where the line crosses the y-axis). It's usually easy to graph a line using y=mx+b. Other forms of linear equations are the standard form and the point-slope form.For example, if you have slope of 2 and points (4, 5) your equation will look like this:5=2x+bif x=4, you get 5=2(4)+bsolve for b: -3y=2x-3
To write the equation 5x + y = 7 in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept, we need to isolate the y variable on one side of the equation. Starting with 5x + y = 7, we subtract 5x from both sides of the equation: y = -5x + 7 Therefore, the equation 5x + y = 7 can be written in slope-intercept form as y = -5x + 7. The slope (m) is -5, and the y-intercept (b) is 7. My recommendation : 卄ㄒㄒ卩丂://山山山.ᗪ丨Ꮆ丨丂ㄒㄖ尺乇24.匚ㄖ爪/尺乇ᗪ丨尺/372576/ᗪㄖ几Ꮆ丂Ҝㄚ07/
The quadratic equation, in its standard form is: ax2 + bx + c = 0 where a, b and c are constants and a is not zero.
ax2 + bx + c = 0
The quadratic formula is used to solve the quadratic equation. Many equations in which the variable is squared can be written as a quadratic equation, and then solved with the quadratic formula.
it is 1
George E. Forsythe has written: 'What is a satisfactory quadratic equation solver?' 'Finite-difference methods for partial differential equations' 'How do you solve a quadratic equation?'
Without an equality sign the given terms of an algebraic expression can't be classed as an equation and so therefore a solution is not possible.
It is: x2-10x+21 = 0 and the value of x is 3 or 7 when solved
It is: 3x2-5x-2 = 0 and the value of x is -1/3 or 2 when solved
There are many ways: one is to factorise. If the quadratic is written as ax2 + bx + c then, if b2 = 4ac, the quadratic is a perfect square. It is (x - b/2a)2
They are page numbers 24 and 25 . ( 24 x 25 = 600 ) The easiest way to solve this is by trial and error. Multiply two consecutive numbers; if the product is too low, try larger numbers, if it is too high, try smaller numbers. You can also write an equation and use the quadratic formula. The equation in this case is x(x+1) = 600. Re-written for use of the quadratic equation, it becomes x2 + x - 600 = 0. This will give you a positive and a negative solution; only the positive solution is sensible in this case.
A quadratic equation is an equation where a quadratic polynomial is equal to zero. It can be written as ax^2+bx+c=0 where a,b,c are the coefficients and x is the variable. A quadratic equation has always two complex solutions for x given by the formula x=-b/2a+sqrt(b^2-4ac)/2a and x=-b/2a-sqrt(b^2-4ac)/2a. Examples of quadratic equations are x^2+x-2=0, 5x^2+6x=0, x^2+1=0 etc.