Answer:
Step-by-step explanation:
if C is the right angle,
AB is the hypotenuse
AB^2=AC^2+BC^2
AC^2=13^2-5^2
AC^2=169-25=144
AC=
=12
AC=12
The required distance would be 17.88 units coordinates A(-4,5) and B(12,13) and the horizontal distance is 16 units and the vertical distance is 8 units from A to B which are determined by the graphing method.
<h3>What is the distance between two points?</h3>
The distance between two points is defined as the length of the line segment between two places representing their distance.
Given AB with coordinates A(-4,5) and B(12,13).
The formula of the distance between two points is A(x₁, y₁) and B(x₂, y₂) is given by: d (A, B) = √ (x₂ – x₁)² + (y₂ – y₁) ².
x₁ = -4, y₁ = 5
x₂ = 12, y₂ = 13
distance = √ (12 – (-4))² + (13 – 5)²
distance = √ (12 + 4)² + (8)²
distance = √ (16)² + (8)²
distance = √ (256 + 64)
distance = √320
distance = 17.88 units
The horizontal distance is 16 units and the vertical distance is 8 units from A to B which are determined by the graphing method.
Learn more about the distance between two points here:
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Answer:
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C
Step-by-step explanation:
∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗
Let 1st=arctan(x)
And 2nd=1
∫▒〖arctan(x).1 dx=arctan(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗
As we know that
derivative of arctan(x)=1/(1+x^2 )
∫▒〖1 dx〗=x
So
∫▒〖arctan(x).1 dx=arctan(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1
Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now
Let 1+x^2=u
du=2xdx
Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get
1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)
1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)
1/2 ∫▒(2xdx/u) =1/2 ln(u)+C
1/2 ∫▒(2xdx/u) =1/2 ln(1+x^2 )+C
Putting values in Eq1 we get
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C (required soultion)
Step 1: Create an equation with a slope of 6
y=6x+b
Step 2: Substitute x and y by with the point (1,2) and solve the equation for b
y=6x+b
2=6(1)
2=6
2=6+b
b=-4
Step 3: Substitute -4 for b in the equation
y=6x+b
y=6x+(-4)
y=6x-4
The equation that has a slope of 6 and passes through the point (1,2) in point-slope form:
y=6x-4