
How do you solve sqrt (6x-5)+10=3? - Socratic
Mar 12, 2017 · √6x − 5+10 −10 = 3 −10 ⇒ √6x −5 = −7 square both sides (√6x − 5)2 = (− 7)2 ⇒ 6x − 5 = 49 add 5 to both sides. 6x−5 +5 = 49+ 5 ⇒ 6x = 54 divide both sides by 6 6x 6 = 54 6 ⇒ …
Question #733f9 + Example - Socratic
Let f (x)=6x^3-7x^2-9x-2 and substitute these values to x and you will find f (2)=0. This means that x=2 is one of the roots and thus f (x) is divisivle by (x-2).
What are the real zeros of this function #f (x)=3x^3+6x^2
f (x) has zeros +-sqrt (5) and -2 Given: f (x) = 3x^3+6x^2-15x-30 Note that all of the terms are divisible by 3. In addition, the ratio of the first and second terms is the same as that of the third …
Question #9d6c7 - Socratic
Let "mass of X" = x and "mass of O" = y. y/ (x + y) = 0.6 y = 0.6x + 0.6y 0.4y = 0.6x y = (0.6x)/0.4 = 1.5x Let "mass of x = 32 u". Then "mass of O = 1.5 × 32 u = 48 u". ulbb ("Element"color …
How do you factor #6x^2+x-1# - Socratic
Jan 29, 2017 · 6x^2+x-1 = (2x+1) (3x-1) Here are a couple of methods (in no particular order): Method 1 Note that: (ax+1) (bx-1) = abx^2+ (b-a)x-1 Comparing with: 6x^2+x-1 we want to find …
Question #80b00 - Socratic
Measure of angle LhatOM = 48^0 7x-92+6x+12=180 1) Combine like terms 7x - 92 + 6x + 12 =180 13x - 80= 180 2) Add color (blue) (80) to both sides (13x) - cancel80 color (blue) (+ …
Find all real zeros of the polynomial function. f (x) = 6x^3
0, 1, 3 To solve this, we set the equation equal to 0 and factor it. First, we can factor out 6x to get 6x(x^2-4x+3)=0. Next we can factor out the area in parentheses. To do this, we need to find …
How do you use the important points to sketch the graph of y
How do you use the important points to sketch the graph of y = x2 − 6x + 1? Algebra Quadratic Equations and Functions Quadratic Functions and Their Graphs
How do you divide # (12x^4-18x^3+36x^2)-:6x^3#? - Socratic
We're dividing a single term, #6x^3#, into a polynomial, so we can take advantage of the following fact which allows us to effectively divide one term into other terms:
Question #9adb8 - Socratic
Explanation: The problem #int (2x)/ (x^2+6x+13)dx# can be solved By the method as shown below. Let #u=x^2+6x+13#, #du= (2x+6)dx# Further, in the numerator, we have only …