0
What else can I help you with?
After being rearranged and simplified which of the following equations could be solved using the quadratic formula?
9x+3x2=14+x-1 AND 2x2+x2+x=30
What two numbers multiply to 280 and add up to 27?
307
What is the constant k in directs and inverse equations?
The constant could be any number.
How do you compute the intercepts of a quadratic function?
Use the quadratic equation. If ax+bx+c=0 x=(-b±(b^2-4ac)^(1/2))/2a. You could also complete the square, factor,or graph the equation.
A rectangle field that is 119 square yards in area is to covered by sod. it is known that the field is 3 yards longer than twice the width. find the dimensions of the field. how do i do this?
It is given that the length is twice the width plus 3 yards; this can be written as a formula: Length = 2 × Width + 3 To calculate the area of a rectangle the formula is: Area_rectangle = Length × Width Substitute for Length in the above formula by the given expression involving Width → Area = Length × Width = (2 × Width + 3) × Width = 2 × Width² + 3 × Width But the area is given as 119 sq yd. If I now write w for Width, this gives: 2w² + 3w = 119 sq yd → 2w² +3w - 119 = 0 Which is a quadratic which can be solved either by using the formula or by factorising: It does factorise: the factor pairs of 119 are 1 × 119 and 7 × 17. One bracket will have (2w ...) and the other will have (w...). As the 119 is negative, there are effectively four pairs of numbers that could be fitted into the brackets: {-1, 119}; {-7, 17}; {7, -17}; {1, 119}. There is one pair which will fit so that when multiplied out gives the quadratic. A quadratic will always have two solutions; as one of the numbers above is negative and one positive, the two solutions will comprise of one positive and one negative value for w (Width); as a length must be positive, the negative solution cannot work and the positive solution is the required Width, from which the Length can be calculated (as above).
After being rearranged and simplified which of the following equations could be solved using the quadratic formula?
9x+3x2=14+x-1 AND 2x2+x2+x=30
If 2x2 plus 5x-12(2x-3)(x plus 4) which equations should be solved to find the roots of 2x2 plus 5x-120?
To find the roots of the equation (2x^2 + 5x - 120 = 0), you first need to set the equation equal to zero. This can be done by solving the quadratic equation directly. Alternatively, you could factor it if possible or use the quadratic formula (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where (a = 2), (b = 5), and (c = -120).
How do students apply quadratic equation as an activity?
Teachers can find many ways to teach students the quadratic equation. An activity could include having contests where students race to solve the equations in the fastest time.
Is it possible to have different quadratic equations with the same solution Why?
Yes it is possible. The solutions for a quadratic equation are the points where the function's graph touch the x-axis. There could be 2 places to that even if the graph looks different.
Are quadratic functions symmetric with respect to the y-axis?
No, but they are symmetric with respect to a line parallel to the y-axis - which could be the y-axis itself.
Why would one use a quadratic equation?
One true purpose for which you could use a quadratic equation could be to figure the maximum use of crop planting in a field once you have the true area of the field figured. Quadratic equations can also be used to figure the area or the volume or the capacity of any unusually shaped object.
What equation could you use to solve o equals 4x2-3x-2?
Quadratic equation formula
What is 5x squared plus 7x plus 2?
The expression (5x^2 + 7x + 2) is a quadratic polynomial in standard form, where (5) is the coefficient of (x^2), (7) is the coefficient of (x), and (2) is the constant term. This polynomial can be used in various mathematical contexts, such as finding roots, graphing, or solving equations. To analyze it further, you could factor it or apply the quadratic formula if you need to find its roots.
What systems of equations could be solved to determine the number of women in the choir?
When you give us a multiple choice question and don't include the choices, we feel sad.
Who found quadratic equation?
The Babylonians, as early as 1800 BC (displayed on Old Babylonian clay tablets) could solve a pair of simultaneous equations of the form: : which are equivalent to the equation:[1] : The original pair of equations were solved as follows: # Form # Form # Form # Form # Find by inspection of the values in (1) and (4).[2] In the Sulba Sutras in ancient India circa 8th century BCE quadratic equations of the form ax2 = c and ax2 + bx = c were explored using geometric methods. Babylonian mathematicians from circa 400 BCE and Chinese mathematicians from circa 200 BCE used the method of completing the square to solve quadratic equations with positive roots, but did not have a general formula. Euclid, the Greek mathematician, produced a more abstract geometrical method around 300 BCE. In 628 CE, Brahmagupta gave the first explicit (although still not completely general) solution of the quadratic equation: : " To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value. (Brahmasphutasiddhanta (Colebrook translation, 1817, page 346)[2] " This is equivalent to: :The Bakhshali Manuscript dated to have been written in India in the 7th century CE contained an algebraic formula for solving quadratic equations, as well as quadratic indeterminate equations (originally of type ax/c = y). Mohammad bin Musa Al-kwarismi (Persia, 9th century) developed a set of formulae that worked for positive solutions. Abraham bar Hiyya Ha-Nasi (also known by the Latin name Savasorda) introduced the complete solution to Europe in his book Liber embadorum in the 12th century. Bhāskara II (1114-1185), an Indian mathematician-astronomer, gave the first general solution to the quadratic equation with two roots.[3] The writing of the Chinese mathematician Yang Hui (1238-1298 AD) represents the first in which quadratic equations with negative coefficients of 'x' appear, although he attributes this to the earlier Liu Yi. The Babylonians, as early as 1800 BC (displayed on Old Babylonian clay tablets) could solve a pair of simultaneous equations of the form: : which are equivalent to the equation:[1] : The original pair of equations were solved as follows: # Form # Form # Form # Form # Find by inspection of the values in (1) and (4).[2] In the Sulba Sutras in ancient India circa 8th century BCE quadratic equations of the form ax2 = c and ax2 + bx = c were explored using geometric methods. Babylonian mathematicians from circa 400 BCE and Chinese mathematicians from circa 200 BCE used the method of completing the square to solve quadratic equations with positive roots, but did not have a general formula. Euclid, the Greek mathematician, produced a more abstract geometrical method around 300 BCE. In 628 CE, Brahmagupta gave the first explicit (although still not completely general) solution of the quadratic equation: : " To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value. (Brahmasphutasiddhanta (Colebrook translation, 1817, page 346)[2] " This is equivalent to: :The Bakhshali Manuscript dated to have been written in India in the 7th century CE contained an algebraic formula for solving quadratic equations, as well as quadratic indeterminate equations (originally of type ax/c = y). Mohammad bin Musa Al-kwarismi (Persia, 9th century) developed a set of formulae that worked for positive solutions. Abraham bar Hiyya Ha-Nasi (also known by the Latin name Savasorda) introduced the complete solution to Europe in his book Liber embadorum in the 12th century. Bhāskara II (1114-1185), an Indian mathematician-astronomer, gave the first general solution to the quadratic equation with two roots.[3] The writing of the Chinese mathematician Yang Hui (1238-1298 AD) represents the first in which quadratic equations with negative coefficients of 'x' appear, although he attributes this to the earlier Liu Yi.
How could Commerce problems be solved?
Commerce problems could be solved with trade
How can yu identify the three variables in an experiment?
This type of experiment is common. For example, which of the elements of the new poultry feed produced the best outcome for the least cost? -- A common industrial task. This is solved by the use of simultaneous equations. The extreme example of a multi-variable experiment is in the cat-scan or similar measurement, where many thousand simultaneous equations are solved in a similar number of variables, to produce the end result. These numeric analyses could never be solved using human calculation power, for the analysis time would be too long. Only with the digital computer, and an algorithm for solving massive simultaneous equations, did this become possible.
