2017年10月4日 · (2y^2+x)^7= 128 y^14+448 x y^12+ 672 x^2 y^10 + 560 x^3 y^8 + 280 x^4 y^6+ 84 x^5 y^4+ 14 x^6 y^2+ x^7 When expanding an expression like this, knowing Pascal's Triangle is crucial: (Source) Since this expression is of degree 7 (i.e. has highest power 7), there will be 8 terms in the expansion (like in a quadratic equation where the highest power is 2, there are 3 …

2017年1月29日 · How do you solve quadratic inequality, graph, and write in interval notation #x^2 - 8x + 15 >0

2018年5月8日 · (The explaining of polynomial long division) Now factorise the quadratic. #2x^2 - 9x+7 = x^2-9x+14# Which factors add up to nine and their product is 14?

2016年3月9日 · How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#?

2018年5月16日 · P(2 < X < 7) = 54.43%. The picture is a bit blurry, but it appears as though X is the number of calls over a 1.5 minute interval.

2017年9月17日 · The zeros of f(x) are 7, i, -i, all with multiplicity 1. Given: f(x) = x^3-7x^2+x-7 Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

2018年3月4日 · x=-12 2x+7= -17 2x= -17-7 2x= -24 (Divide both sides by 2) x=-12. Algebra Linear Equations Two-Step Equations and Properties of Equality

2015年5月2日 · #2x - 8 < 7# Adding 8 to both sides will give us: #2x - cancel(8) + cancel(8) < 7 + 8# #2x < 15# Dividing both sides by 2 will give us: #(cancel(2)x)/cancel(2) < 15/2# #x < 15/2# #color(green)(x<7.5# To verify your answer, choose an appropriate value of #x# and see if the inequality is satisfied. Say #x = 2# (A random number less than 7.5)

#2x+2*7+x=20# Multiply by #7# to get #14# #2x+14+x=20# Add #2x# and #x# to get #3x# #3x+14=20# Since #14# does not contain the variable to solve for, move it to the right side of the equation by subtracting #14# from both sides. #3x=-14+20# Add #-14+20# to get #6# #3x=6# Divide each term by #3# and simplify. Answer : #x=2#

2016年8月3日 · In interval notation, this is written as x in [5, +oo) For the second condition, you need x to be smaller than or equal to 18. In interval notation, this is written as x in (-oo, 18] This means that the solution interval for the compound inequality must have x greater than or equal to 5 and smaller than or equal to 18.