3 = 4 - 4^0 * 4/4
4 = 4^0 + 4^0 + 4^0 + 4^0
5 = 4^2 / (4^0 * 4) + 4^0
6 = 4^2 / 4 + 4^0 + 4^0
All points that lie in a given line with a defined equation, they satisfy the equation such that when the values of x and y are substituted they satisfy the given equation.
Substituting values of y and x in the equation y = 14x +4
(8,6), (-8,-5), (-16,0), (-20,1) doesn't satisfy the equation.
therefore in this case there is no point that would lie in the line
Answer:
8:16
Step-by-step explanation:
Chapter : Algebra
Study : Math in Junior high school
x = 7 + √40
find √x of √x + 1
= √x + 1
= √(7+√40) + 1
in Formula is :
= √7+√40 = √x + √y
= (√7+√40)² = (√x + √y)²
= 7+√40 = x + 2√xy + y
= 7 + √40 = x + y + 2√xy
→ 7 = x + y → y = 7 - x ... Equation 1
→ √40 = 2√xy → √40 = 2.2√10 = 4√10
= xy = 10 ... Equation 2
substitution Equation 1 to 2 :
= xy = 10
= x(7-x) = 10
= 7x - x² = 10
= x² - 7x + 10 = 0
= (x - 5)(x - 2) = 0
= x = 5 or x = 2
Subsitution x = 5 and x = 2, to equation 1
#For x = 5
= y = 7 - x
= y = 7 - (5)
= y = 2
#For x = 2
= y = 7 - x
= y = 7 - (2)
= y = 5
and his x and y was find :
#Equation 1 :
= x = 5 and y = 2
#Equation 2 :
= x = 2 and y = 5
So that :
√7+√40 = √x + √y
= √7+√40 = √2 + √5
And that is answer of question :
= √2 + √5 + 1
Answer:
see below for a graph
Step-by-step explanation:
Completing the square for x and y terms can help you put this in the standard form.
(x^2 +10x) +(y^2 +6y) = -18
(x^2 +10x +25) +(y^2 +6x +9) = -18 +25 +9 . . . . add square of half linear coefficient
(x +5)^2 +(y +3)^2 = 16 . . . . . . . circle centered at (-5, -3) with radius 4
