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# I have a math test tomorrow and i need help!? Topic: I have a math test tomorrow and i need help!?
July 18, 2019 / By Arnold
Question: Ok so I have a math test tomorrow. It is on slopes, rate of change, direct and partial variation, equations, etc. I understand most of it although there are some things that I still do not understand. There was this one question and it was asking you to find the coordinates of a possible end point and it gave you the coordinates for the first point, and the slope. The first coordinates were: (5,-3) and the slope was: -4. If the slope was a fraction, I would know how to solve this problem although I do not know how to make a slope into a fraction if it already isn't one. Another thing I have a little bit of a problem with is writing an equation for a graph AND making a graph based on an equation. Please help?! Thank you! ## Best Answers: I have a math test tomorrow and i need help!? Una | 7 days ago
y = mx+b -3 = -4(5)+b -3 = -20+b -3+20 = b 17 = b Y = -4x+17 All I did was substitute the order pair and the slope in y = mx+b and solve for b. To graph this all you do is plug value in for x and solve for y Let x = 0 y = 17 (0,17) is on your line Let x = 1 y = -4(1)+17 y = -4+17 y = 13 (1,13) is on the line as well Blessings
👍 262 | 👎 7
Did you like the answer? I have a math test tomorrow and i need help!? Share with your friends Originally Answered: PLEASE HELP ME! HUGE MATH TEST TOMORROW!?
Okay - the key to doing ALL of these problems is to remember that you have to keep your apples, your oranges, and your elephants all SEPARATE. You can't combine p and k or t and u. You can't combine z with plain old numbers. So - let's go through the first one together, and you can work on the rest: 12p = 6 - 3p Now - the only thing you can do to an equation is the SAME THING to BOTH SIDES. Just like a scale - they have to keep the same value. So - how can you group the "p"s together. Well - add 3p to both sides: 12p + 3p = 6 - 3p + 3p 15 p = 6 {and the p's go away! 0p = 0} Now, divide both sides by 15: 15p / 15 = 6 /15 p = 6/15 = 2/5 This is how you do all of these problems. Good luck! Saranna
The slope is a fraction. It is (-4)/1 or 4/(-1). Now the harder questions. Writing the equation of the graph: Take all the points that cross the x axis (the roots). Then write them in this form: (x-a), (x+b), (x-c), ....... Then just multiply them together (x-a)*(x+b)*(x-c) Drawing the graph of an equation: You can plot points. I don't know your level of Math experience, but here is where calculus can help. Find the roots of the equation (opposite approach mentioned above). They will be of this form (a,0), (-b,0), (c,0), .... If you take the derivative of the equation and solve for x, you will get the x points that correspond to the maxima and minima of the graph. plug these x values into the original equation and solve for y. These coordinates will give the max and min points of the graph. The max will be the one with the largest y value and the min will be the one with the smallest y value. If you take the second derivative of the original equation and solve for x and then y (from the original), this will give the flex points. These are where the graph changes from and increasing to a decreasing function or vice versa. Once you have the roots, the max/min and the flex points identified on a graph, you should be able to draw the entire graph. This is probably more than you wanted to know.
👍 110 | 👎 1 Nonie
If you have an initial point at (5,-3) and the slope is -4 then some points on the line would be (6,-7), (4,1) or (3,5). If you are given a whole number as a slope to put it in fraction form just put a 1 in a denominator (In this case it is -4/1) This gives you a RISE/RUN. To make the equation in y=mx+b you are given slope so all you have to find is the y-intercept. Based on the information you gave me it is 17 because you rise 4 units per one unit left. To see a graph, go to www.wolframalpha.com and plug in the equation y=-4x+17.
👍 104 | 👎 -5 Mab
when the slope is a whole number, like -4 you need to look at as -4/1 two ways to graph solve for x=0 and y=0 gives you (0,y) and (x,0) plot those points and connect the dots if the values are outrageous pick one I prefer y=0 gives you (x,0) then use the slope to generate in-between points if the slope is -4 y= -4x slope= -4x/1 if you have y= -4x+412 when y=0, x=103 your point is (103,0) draw a line [horizontal LOL] plop (103,0) on the line go down 4 and 1 to the right or 4 up and 1 to the left draw your line through those two points good luck
👍 98 | 👎 -11 Kelia
If your slope is -4, just write it over 1. There is your fraction. -4/1 so the difference between the two y values is -4. The difference between the two x values is 1.
👍 92 | 👎 -17 Irene
Assume (5, - 3) and slope = - 2/5 - 2/5 = (y - ( - 3))/(x - 5) - (2/5)*(x - 5) = y + 3 - (2/5)/x + 2 = y + 3 y = - (2/5)*x - 1
👍 86 | 👎 -23 Originally Answered: I need help on two math problems that I need to know for my test tomorrow?
For the first question since the equation is a quadratic the 2 roots will be the complex number you have and its conjugate. Since the coefficient of x^2 is 1 you can say that the product of the roots will give the -b, where b is the coefficient of x and the sum of roots will give you c. Notice that on multiply and adding a complex number to its conjugate you get real numbers. If x^2 had a coefficient say, a, the the formulae would become product of roots = -b/a sum of roots = c/a The conjugate of a complex is number is the number with the sign of the imaginary part reversed. For the second question you can differentiate the function with respect to time and then equate it to 0. This gives you the value for t at which there exists either a maxima, minima or point of inflection. Since it is obvious from the question that you could only have got the maxima simply substitute the value for t that you get into the function to get that maximum value of the function. If you are not sure if you have obtained a maxima or minima get the double derivative of t and then check the sign of the function on substituting the value of t obtained earlier. If it is positive you have a minima for that value of t, if it is negative you have a maxima.

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