2231 Shares

HELP! Did i do this right? Algebra problem solved by using quadratic applications?

HELP! Did i do this right? Algebra problem solved by using quadratic applications? Topic: Problems of problem solving method
June 20, 2019 / By Doreen
Question: I dont think anyone would actually like to work this out and check but if u do thanks so much!!!!! Here's the problem: From 1990-2003 the amount of noodles (y) imported to the US can be modeled by the function y= 1.36X^2+27.8X+ 304 where X is the number of years since 1990. Write and solve an equation that you can approximate the year when 500 million pounds of noodles were imported. For the equation I got 500= 1.36x^2+27.8x+304 and for the year i got 1995 (maybe 1996 i'm not sure if i should round up) Is this correct? If not than what is? THANKS!
Best Answer

Best Answers: HELP! Did i do this right? Algebra problem solved by using quadratic applications?

Cassiah Cassiah | 1 day ago
Your answer is quite correct. The equation is indeed 500 = 1.36x² + 27.8x + 304, or 136x² + 2780x – 19600 = 0; whence, solving by your method of choice, x ≈ 5.5457706 or –25.98694707. I suggest rounding up the positive root; whence, x ≈ 6, giving us the year 1996. I assume that y is intended to be measured in millions of pounds. I hope that this answer helps you.
👍 210 | 👎 1
Did you like the answer? HELP! Did i do this right? Algebra problem solved by using quadratic applications? Share with your friends

We found more questions related to the topic: Problems of problem solving method


Cassiah Originally Answered: How is this algebra 1 problem solved?
Hi, Hope I can help, First, you need to get rid of the parentheses by distribution (3(x-4)=2y) = (3(x)-3(4)=2y) = (3x-12=2y) (3x-12=2y), you can now move the "2y" to the left side (3x-12=2y) = (3x-12-2y =2y - 2y) = (3x-12-2y =0), now you can move the (-12) to the right side (3x-12-2y =0) = (3x-12+ 12 -2y =0 + 12) = (3x - 2y = 12), this is the first equation in standard form Now lets do the second equation (6(y+6)=9x), using distribution (6(y+6)=9x) = (6(y)+6(6)=9x) = (6y+36)=9x) Now we can reduce the equation by dividing each side by "3" (6y+36)=9x) = ((6y+36)/3=(9x/3)) =( (6y/3)+(36/3)=9x/3) = ( (2y)+(12)=(3x)) ( (2y)+(12)=(3x)) = ( 2y+12=3x), now move the "2y" to the right side ( 2y+12=3x) = ( 2y -2y+12=3x - 2y) = ( 12=3x - 2y) or ( 3x - 2y =12) ( 3x - 2y =12) is the second equation . . . The two equations in reduced form are (1) (3x - 2y = 12) (2) (3x - 2y = 12) As you can see, these two equations are the same, meaning, that these two equations are the same line. They would have an infinite amount of solutions, if this was a 2 system equation If you wanted to plot a graph of this, all you do is replace "x" with ANY number Lets replace "x" with "0" (3x - 2y = 12) = (3(0) - 2y = 12), now just solve for "y" (3(0) - 2y = 12) = ( 0 - 2y = 12) = ( - 2y = 12), now divide each side by (-2) ( - 2y = 12) = ( - 2y / -2 = 12 / -2 ) = ( y = -6 ), you can check your answer by replacing "x" with "0" and "y" with (-6) in the equation (3x - 2y = 12) = (3(0) - 2(-6) = 12) = ( 0 + 12 = 12) = (12 = 12) (True) x = 0 y = -6 points, or ordered pairs are given as ( x,y ) , so this point would be ( 0,-6 ) to find again start from the beginning Replace "x" with any number, lets replace "x" with (-4) (3x - 2y = 12) = (3(-4) - 2y = 12) = ((-12) - 2y = 12) solve for "y" move the (-12) to the right side ((-12) - 2y = 12) = ((-12) + 12 - 2y = 12 + 12 ) = ( - 2y = 24), now divide each side by (-2) ( - 2y = 24) = ( - 2y / -2 = 24/ -2) = ( y = -12 ), now check by replacing "x" with (-4), and "y" with (-12) (3x - 2y = 12) = (3(-4) - 2(-12) = 12) = ( -12 + 24 = 12) = ( 12 = 12 ) (True) x = ( -4 ) y = ( -12 ) Points are given as (x,y), our point is ( -4, -12 ) Now draw a line between the two points ( 0, -6 ) and ( -4, -12 ), and that will be the graph . If you want to find more points you can Just follow the steps (1) replace "x" with any number, (3x - 2y = 12) = (3(8) - 2y = 12) (2) solve for "y", (3(8) - 2y = 12) = ( 24 - 2y = 12) = (24 - 24 - 2y = 12 - 24) = ( - 2y = -12 ) = ( - 2y / -2 = -12 / -2 ) = ( y = 6 ) (3) Check your answer by replacing "x" and "y" with their numbers, (3x - 2y = 12) = (3(8) - 2(6) = 12) = (24 - 12 = 12) = ( 12 = 12 ) (True) (4) Put numbers in point form, (x,y) ( 8,6 ) (8,6) is another point You can try algebra.com, a free tutoring site, where you can ask any math question, and get 1-10+ answers, they would also be able to graph this for you, and give easy to read formulas Hope I helped
Cassiah Originally Answered: How is this algebra 1 problem solved?
Plots can be anything you want, just make sure you do your math correctly. This is easier if you put these equations into y-intercept form. Both of these equations become. y=(3/2)x-6 so now you can plug any number into x to see what y is. so when x is 0, y is -6. When x is 1 y is -9/2 When x is 2, y is -3 When x is 3 y is -3/2 etc. When you graph this, it comes out as a straight, upward sloping line. As for your question about possibility, I do not understand what you are trying to prove possible, there is information missing. There are many possibilities to what I am proving. For instance, I just proved that they are the same line, and I also just proved that they are lines. I also proved that they are not parallel, or perpendicular.
Cassiah Originally Answered: How is this algebra 1 problem solved?
y=mx+b m=rise/run 1. Multiply a term by terms in the parentheses. 2. Solve for y. 3. Graph (0,b) 4. Plot new point by following the slope formula.

Anna Anna
let the negative number you are looking for be represented by "a" then, you can set up an equation based off the info given: (-a+3)^2=1521 Solve for a (-a+3)= +/- Sqrt(1521) -a= +/- Sqrt(1521) get rid of the negative sign on the left by multiplying by -1 So, a = -Sqrt(1521)+3 or a =+Sqrt(1521)-3 But the question asks for a negative real number, so the answer for this question is a= -Sqrt(1521)+3= -36
👍 90 | 👎 -8

Anna Originally Answered: Algebra problem involving quadratic stuff- Help?
The discriminant is defined as b^2 - 4ac. Meaning you plug in a, b, and c from the quadratic equation and simplify it. Once you have the discriminant you can determine ho much solutions the quadratic equation has. If the discriminant is greater than 0, the equation has 2 real solutions. If the discriminant is equal to 0, the equation has 1 real solution. Finally, if the discriminant is less than 0, you have 2 unreal solutions. The quadratic formula is below. Edit - In case you have trouble understanding me, http://www.purplemath.com/modules/quadform2.htm x = -b ± sqrt(b^2-4ac)/2a Just remember, a formula needs the values plugged in for them to work. Meaning for the quadratic formula, and the formula to find the discriminant you need to plug in a, b, and c. If we had the equation 2x^2 - 3x + 4. a = 2, b = -3, c = 4.
Anna Originally Answered: Algebra problem involving quadratic stuff- Help?
a) In the quadratic equation x=(-b±sqrt(b^2-4ac))/2a, the discriminant is the b^2-4ac part. Just substitute the values in. b) The discriminant describes how many roots an equation will have. This means the number of values of x for which y=0; the solutions. If the discriminant is negative, there are none. If it is zero, the graph just touches the x axis, so there is one solution. If it is positive, there are two solutions. c) If the discriminant is greater than or equal to zero, just substitute the values into that equation to get your solution(s) for x

If you have your own answer to the question problems of problem solving method, then you can write your own version, using the form below for an extended answer.