Answer:
39 ft²
Step-by-step explanation:
Area of a triangle = 1/2bh
Let:
base = b = 4 ft
height = h = 3 ft
Solve for the area of the triangle:
A = 1/2(4)(3)
A = 2 * 3
A = 6 ft²
Area of a rectangle = bh
Let:
base = b = 9 ft
height = h = 5 ft
Solve for the area of the rectangle:
A = (9)(5)
A = 9 * 5
A = 45 ft²
Subtract the area of the triangle from the area of the rectangle:
45 - 6 = 39 ft²
39 ft² is your answer.
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Answer:
5 10
2 4
Step-by-step explanation:
To multiply these expression, multiply the numeric parts, and sum the exponents of the literal parts (because the variable is the same: we're using the power property 
).
So, you have:
Answer:
The area of the square adjacent to the third side of the triangle is 11 units²
Step-by-step explanation:
We are given the area of two squares, one being 33 units² the other 44 units². A square is present with all sides being equal, and hence the length of the square present with an area of 33 units² say, should be x² = 33 - if x = the length of one side. Let's make it so that this side belongs to the side of the triangle, to our convenience,
x² = 33,
x = 
.... this is the length of the square, but also a leg of the triangle. Let's calculate the length of the square present with an area of 44 units². This would also be the hypotenuse of the triangle.
x² = 44,
x = 
.... applying pythagorean theorem we should receive the length of a side of the unknown square area. By taking this length to the power of two, we can calculate the square's area, and hence get our solution.
Let x = the length of the side of the unknown square's area -
= 
+ 
,
x = 
... And squared is 11, making the area of this square 11 units².
Given three points
P1(-2,8)
P2(0,-4)
P3(4,68)
We need the quadratic equation that passes through all three points.
Solution:
We first assume the final equation to be
f(x)=ax^2+bx+c .............................(0)
Observations:
1. Points are not symmetric, so cannot find vertex visually.
2. Using the point (0,-4) we substitute x=0 into f(x) to get
f(0)=0+0+c=-4, hence c=-4.
3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.
f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)
f(4)=a(4^2)+b(4)-4=68 => 16a+4b-4=68.............(2)
4. Solve system
2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)
substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2 ..........(4)
5. Put values c=-4, a=4, b=2 into equation (0) to get
f(x)=4x^2+2x-4
Check:
f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8
f(0)=0+0-4 = -4
f(4)=4(4^2)+2(4)-4=64+8-4=68
So all consistent, => solution ok.
