The corresponding height of the triangle is 1.6 units
The formula for calculating the area of a triangle is expressed as:
- b is the base of the triangle
- h is the height of the triangle
Given the coordinates of the base BC of the triangle given as B(3, 2), and C(-1,-1). Using the distance formula:
The area of the triangle passing through the coordinate points A(-1, 1), B(3,2), and C(-1, -1) is expressed as:
![A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]\\](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5Bx_1%28y_2-y_3%29%2Bx_2%28y_3-y_1%29%2Bx_3%28y_1-y_2%29%5D%5C%5C)
Substituting the coordinate points:
![A=\frac{1}{2}[(-1)(2-(-1))+3(-1-1)+-1(1-2)]\\A=\frac{1}{2}[(-1)(3)+3(-2)+-1(-1)]\\A=\frac{1}{2}[-3-6+1]\\A=\frac{1}{2} (-8)\\|A| =4 units^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5B%28-1%29%282-%28-1%29%29%2B3%28-1-1%29%2B-1%281-2%29%5D%5C%5CA%3D%5Cfrac%7B1%7D%7B2%7D%5B%28-1%29%283%29%2B3%28-2%29%2B-1%28-1%29%5D%5C%5CA%3D%5Cfrac%7B1%7D%7B2%7D%5B-3-6%2B1%5D%5C%5CA%3D%5Cfrac%7B1%7D%7B2%7D%20%28-8%29%5C%5C%7CA%7C%20%3D4%20units%5E2)
Recall that:
Hence the corresponding height of the triangle is 1.6 units
Learn more on area of triangles here: brainly.com/question/17335144
In this question, the thing that needs to be found is the number whose 20% would be 18. The information's required for getting to the solution are already provided in the question. based on those given information's, the answer can be deduced very easily.
Let us assume the unknown number to be = x
Then
(20/100) * x = 18
x/5 = 18
x = 18 * 5
= 90
so the unknown number in the question is 90. I hope the procedure is clear enough for you to understand.
Answer:
-4/5
Step-by-step explanation:
To find the slope of the tangent to the equation at any point we must differentiate the equation.
x^3y+y^2-x^2=5
3x^2y+x^3y'+2yy'-2x=0
Gather terms with y' on one side and terms without on opposing side.
x^3y'+2yy'=2x-3x^2y
Factor left side
y'(x^3+2y)=2x-3x^2y
Divide both sides by (x^3+2y)
y'=(2x-3x^2y)/(x^3+2y)
y' is the slope any tangent to the given equation at point (x,y).
Plug in (2,1):
y'=(2(2)-3(2)^2(1))/((2)^3+2(1))
Simplify:
y'=(4-12)/(8+2)
y'=-8/10
y'=-4/5
46 = 5k - 59
46 + 59 = 5k
105 = 5k
105/5 = k
21 = k
k is 21
