(7x^2-xy+y^2)+(-x^2-8xy+5y^2 can someone please help me solve this, I am drawing a blank.?
First of all, to help, this is an addition problem. As such, you can get rid of the + sign and just reduce it so that you're combining like terms. 7x^2 - xy + y^2 + -x^2 - 8xy + 5y^2 Then, pair the terms with the same variable and exponent and combine. 7x^2 + -x^2 = 6x^2 -xy + -8xy = -9xy y^2 + 5y^2 = 6y^2 (Keep in mind that no coefficient or a negative sign in front of a variable correspond to 1 or -1, respectively.) Thus, the answer is: 6x^2 - 9xy + 6y^2. P.S. Does your teacher want you to put the answer in standard form? If so, I'll add on. I don't want to put anymore info because it might cause you to get the wrong answer.
Can someone help me solve the system of linear equations using substitution?
Equation 1 is y = 5x - 1/2 Substitute for y in Equation 2 6x = (5x - 1/2) + 2 1/2 6x = 5x - 1/2 + 2 12 x = 2 Now substitute for x in the second equation. 6(2) = y +2 1/2 12 = y + 2 1/2 9 1/2 = y Substitute for both x and y in Equation1 to check 9 1/2 = 5(2) - 1/2 9 1/2 = 10 - 1/2 9 1/2 = 9 1/2 Checks!
Given equation is,(x^2-x) y^2+y-(x^2+x)It is Quadratic equation in y of form a*y^2+b*y+cHere,a= x(x-1) , b= 1 , c= x(x+1); where a!=0Solutions to this equation are,y= [-b+(b^2 - 4ac)]/2a , y= [-b-(b^2 - 4ac)]/2a ;Thus, Substituting values,b^2–4ac= 1- (4x^2)*(x^2–1) -b+ b^2–4ac = (4x^2)(x^2–1)y= (4x^2)(x^2–1)/x(x-1)y= 4x(x+1)Now,Product of roots is c/ac/a= x(x+1)/x(x-1) = (x+1)/(x-1) ;Thus, y*4x(x-1) = (x+1)/(x-1) ;y= (x+1)/ [4x(x-1)]Thus, factors of this equation are,y= 4x(x+1) and y= (x+1)/ [4x(x-1)]List of Formulas:1. a^2 - b^2 = (a+b)(a-b)2. ab-ac = a(b-c)
I recommend you don't. Calm down. Try to find a good tutor.If your tutors are not helping you, reading books at you... well, they're awful tutors. Maybe the source? I was a stats tutor in grad school, and I tutored becausea) it helped with the bills; andb) it helped me understand my own studies better.Any tutor who really wants to help his/her student must learn to think in multiple ways, because it is rare that people think about things in the exact same way. It is common, in fact, that two students (say, of mine) will grasp things in completely different ways. One will need a certain emphasis, while the other will look at things completely differently.Unfortunately, if you're studying anything else but statistics, at least from my students' experiences, you're likely being taught the material poorly. There seems to be a bias in statisticians to design classes for non-statisticians that just don't work very well.I strongly recommend you try and find a better tutor. Even if you find some person willing to do your homework for you, unless you posted under a pseudonym, you can get in trouble. Also, the plain fact is, statistics is becoming more important in almost every field. If you don't get it now, even if you keep your 4.0, you'll probably be in trouble later when you need to understand.I feel for you. Some of my students have told me horror stories about previous tutors. Try to find a better one, real or virtual. (You may be able to find a really good tutor whom you can Skype with or something. It's possible. And, if you're wondering, no, I don't tutor anymore.)Another suggestion I can give, if you're having problems with the logic of stats, is to Google it. There are well-taught stats courses that don't assume too much background, and more of them are appearing online. Even reading different descriptions of the areas you're having trouble with might help... although you might need a bit of a math background to do so effectively.)I hope you find what you're looking for, and that "thing" is NOT getting someone to do your homework for you.
Consider the curve defined by 2y^3 + 6x^2y - 12x^2 + 6y = 1 (2 part question!)?
I'm in calculus a/b this year and I'm doing review for my finals, and I can't figure out how to do the rest of this problem. Please help! I already did implicit differentiation and ended up getting: dy/dx = (4x-2xy)/(x^2+y^2+1) and I know for sure that this is the right answer. But can someone help me with the other three parts? b) Write an equation of each horizontal tangent line to the curve. c) The line through the origin with the slope -1 is tangent to the curve at point P. Find the x- and y-coordinates of P. I want a good grade on my finals, so I'd appreciate the help!
Algebra 2 help, pls explain how and why?
The symbols f(x) and g(x) are like names given to relations between two quantities. In Algebra, we are used to looking at equations where y is oftentimes alone on the left side of the equation while an expression involving x is on the right side of the equation. For example, y = 15x - 3(x + 1). This equation tells us that there is a relation between y and x and that relation is expressed by the equation. If you assign a value to x then you can calculate the value of y. If you can assign any value to x and get a sensible value for y then all those values you can assign for x is called the Domain of the function while all those values of y that you get is called the Range of the function. I said function, because when the relation between two quantities is such that for each value of x there corresponds one and only one value of y, then the relation is called a FUNCTION. So, my example, y = 15x - 3(x + 1), is a function. I can call it h(x), p(x), t(x), etc. It tells me that in my function, x is the independent variable, while the function name itself replaces y and acts as the dependent variable. Real numbers can be added, subtracted, multiplied and divided. Functions can also be added, subtracted, multiplied and divided like real numbers. This is exactly what happens in your two functions: f(x) and g(x). They are added, f(x) + g(x), and subtracted, f(g) - g(x). Addition and subtraction of two functions result in the creation of another function, just as another real number is created when you add or subtract two real numbers. Hence, if you look at the sum of f(x) + g(x) which is -5x^1/2, it is quite clear that the values of x for which -5x^1/2 will give you real numbers answers are from 0 and all positive real numbers. The expression x^1/2 will be imaginary number if x is negative. Hence, the DOMAIN for f(x) + g(x) is (0, + ∞). The same is true for f(x) - g(x). Hope I help teddy bo6y
Please help calculus question 6x + y^2 = 16, x = y find the area of the region from y = -8 to 2?
Firstly make a rough sketch of our problem : 6x + y^2 = 16 is a sideways parabola, opening to the left y = x is a 45° line through the origin. Firstly : solve for the intersection points : 6x + y^2 = 16 gives, at y = x the equation 6x + x^2 = 16 or x^2 +6x -16 =0 , factoring to give (x-2)(x+8) = 0 or roots of +2 and -8 ( Hmmm also our integration limits... how nice !!! so the points of intersection are ( 2,2) and -8,-8) Secondly : look at the labeled sketch Deciding on which direction to integrate : We already have the y limits, and Its going to be easier to integrasts a y^2 than a √x form so lets integrate sideways, along the y axis Thirdly , the Area form : our little slice of sideways area is dA = (top - bottom ) dy The Top will be the curve x = (16-y^2)/6 and the Bottom will be the line x = y Sketch our little area element and then dA = [ (16-y^2)/6 -y] dy = [16/6 -y^2/6 -y]dy and with limits of -8 and +2 A = ∫dA ........2 A = ∫16/6 -y^2/6 -y]dy ......-8 .................................2 =(16/6 y -y^3/18 -y^2/2] .................................-8 = 250/9 = 27 7/9 ... Eventually so the exact area is 27 7/9 <----- answer -------------------------- checking on a TI-84 "MATH 9" gave : fnInt(((16-y^2)/6 -y), y, -8,2)= 27.7777777 as expected
for numeratorx^3+3*x^2–6*x+2note the sum of coefficient of x in this polynomial =01+3–6+2=0so x=1 is the factor of this polynomialnow write the above polynomial like i am writting belowx^3+3*x^2–6*x+2=x^2*(x-1)+4*x(x-1)-2*(x-1)x^3+3*x^2–6*x+2=(x-1)(x^2+4*x-2)now for denominatorx^3+3*x^2–3*x-1here also the sum of coefficient of x in this polynomial =0so x=1is the factor of above polynomial so i can write like i have done in for numerator polynomialx^3+3*x^2–3*x-1=x^2*(x-1)+4*x(x-1)+1(x-1)x^3+3*x^2–3*x-1=(x-1)(x^2+4*x+1)so now (x^3+3*x^2–6*x+2)/x^3+3*x^2–3*x-1=(x-1)(x^2+4*x-2)/(x-1)(x^2+4*x+1)=(x^2+4*x-2)/(x^2+4*x+1)now on just putting x=1 in above we found the limit when x approaches x=1so answer=(1^2+4*1–2)/(1^2+4*1+1)=3/6=1/2=0.5
Solve this math equation 6x(y^2z) x=1/2, y=-1, z=2?
bear in ideas PEMDAS. it relatively is fundamental in determining a thank you to simplify equations... (and additionally remembering which you would be able to in straightforward terms combine like numbers (numbers with numbers, variables with variables) Parenthesis Exponents Multiply Divide upload Subtract question a million: 4x+8=3(x+8) Simplify the parenthesis: 4x+8 = (3*x)+(3*8) 4x+8 = 3x+24 4x = 3x+24-8 (subtract 8 from the two facets) 4x = 3x+sixteen 4x - 3x = sixteen (subtract 3x from the two facets) x = sixteen question 2: 2y+ x/2 = 7 - considering that is a linear equation there will be lots of solutions. the least confusing thank you to is to make one variable a 0 and remedy for the equation. One achieveable answer could be: 2y + 0 = 7 y = 7/2 y = 3.5 The ordered pair could be (0,3.5) fixing it any opposite direction could provide an ordered pair of (14,0) question 3. -9z = 7(3+2w)+ -8w+17z+-11 first ingredient of do is take away the + sign before the minus indicators. (do not try this in case you have a +- in front on a pair of parenthesis) -9z = 7(3+2w)-8w+17z-11 -9z = 21+14w-8w+17z-11 (assorted out interior the path of the parenthesis) -9z = 21-6w+17z-11 (combine the w variable) 0 = 21-6w+26z-11 (comebine the z variable) 0 = 26z-6w+10 (simplified) 26z-6w = -10 (simplified into an equation) 13z-3w = -5 (divide the completed equation by employing 2) - maximum individuals will ignore this step. desire this permits.
First of all, let's talk about the numerical values in the given sequenceThe numbers in the given sequence are 6,7,27 is a prime number and hence the next common factors are 14,21 which are not part of the given numericals in the sequence.The above sequence hence only contains x*y as the factorThe sequence can be written asxy{(6x^2)+(7xy)-(20y^2)}Now for the remaining quadratic function, we can use the normal factoration technique.It can be written as(6x^2)+(15xy-8xy)-(20y^2)On further rearranging, it can be written as(6x^2)-8xy +15xy-(20y^2)2x(3x-4y)+5y(3x-4y)(2x+5y)(3x-4y)So the given expression can be written asxy(2x+5y)(3x-4y)