Option C:
Area of the remaining paper = (3x – 4)(3x + 4) square centimeter
Solution:
Area of the square paper =
sq. cm
Area of the square corner removed = 16 sq. cm
Let us find the area of the remaining paper.
Area of the remaining paper = Area of the square paper – Area of the corner
Area of the remaining = ![9x^2-16](https://tex.z-dn.net/?f=9x%5E2-16)
= ![(3x)^2-(4)^2](https://tex.z-dn.net/?f=%283x%29%5E2-%284%29%5E2)
Using algebraic formula: ![a^2-b^2=(a+b)(a-b)](https://tex.z-dn.net/?f=a%5E2-b%5E2%3D%28a%2Bb%29%28a-b%29)
![=(3x-4)(3x+4)](https://tex.z-dn.net/?f=%3D%283x-4%29%283x%2B4%29)
Area of the remaining paper = (3x – 4)(3x + 4) square centimeter
Hence (3x – 4)(3x + 4) represents area of the remaining paper in square centimeters.
Answer:
Center: (4,8)
Radius: 2.5
Equation: ![(x-4)^2+(y-8)^2=6.25](https://tex.z-dn.net/?f=%28x-4%29%5E2%2B%28y-8%29%5E2%3D6.25)
Step-by-step explanation:
It was given that; the endpoints of the longest chord on a circle are (4, 5.5) and (4, 10.5).
Note that the longest chord is the diameter;
The midpoint of the ends of the diameter gives us the center;
Use the midpoint formula;
![(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )](https://tex.z-dn.net/?f=%28%5Cfrac%7Bx_1%2Bx_2%7D%7B2%7D%2C%5Cfrac%7By_1%2By_2%7D%7B2%7D%20%29)
The center is at; ![(\frac{4+4)}{2} ,\frac{5.5+10.5}{2}=(4,8)](https://tex.z-dn.net/?f=%28%5Cfrac%7B4%2B4%29%7D%7B2%7D%20%2C%5Cfrac%7B5.5%2B10.5%7D%7B2%7D%3D%284%2C8%29)
To find the radius, use the distance formula to find the distance from the center to one of the endpoints.
The distance formula is;
![d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
![r=\sqrt{(4-4)^2+(10.5-8)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B%284-4%29%5E2%2B%2810.5-8%29%5E2%7D)
![r=\sqrt{0^2+(2.5)^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B0%5E2%2B%282.5%29%5E2%7D)
![r=\sqrt{0^2+(2.5)^2}=2.5](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B0%5E2%2B%282.5%29%5E2%7D%3D2.5)
The equation of the circle in standard form is given by;
![(x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
We substitute the center and the radius into the formula to get;
![(x-4)^2+(y-8)^2=2.5^2](https://tex.z-dn.net/?f=%28x-4%29%5E2%2B%28y-8%29%5E2%3D2.5%5E2)
![(x-4)^2+(y-8)^2=6.25](https://tex.z-dn.net/?f=%28x-4%29%5E2%2B%28y-8%29%5E2%3D6.25)
Whole formula 2*pi*radius^2 + 2*pi*radius*height:
157+942= 1099
SA= 1099
Answer:
y=3x-7
Step-by-step explanation:
lines are parallel hence gradient from the equation in question is the same as the gradient of the equation to be found.. comparing to y=mx+c, eq in question has grad 3... from the formula y-y1=m(x-x1) where (x1,y1) is equal to the point in question
-3x-12> 8x + 21
+12 +12
-3x > 8x + 33
-8x -8x
-11x > 33
————-
-11
x < -3
I moved the real numbers to one side and and the variable to the other side to solve for the inequality
I also remembered somewhere that if you divide by a negative number the inequality sign switches.