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:
4
Step-by-step explanation:
If Nicholas has 14 games, Daqauvion has 11 games because 14-3=11
Answer:
Area = 3.3998
Perimeter = 2.9
Step-by-step explanation:
A method for calculating the area of a triangle when you know the lengths of all three sides.
Let a,b,c be the lengths of the sides of a triangle. The area is given by:
Area = √ p ( p − a ) ( p − b ) ( p − c )
where p is half the perimeter, or
a + b + c / 2
p = 1.7 + 1.7 + 2.4 / 2 = 5.8 / 2 = 2.9
a = Area = 3.3998
Heron was one of the great mathematicians of antiquity and came up with this formula sometime in the first century BC, although it may have been known earlier. He also extended it to the area of quadrilaterals and higher-order polygons.
