<u>Given:</u>
The equation of the line passes through the point (6,9) and is perpendicular to the line whose equation is 
We need to determine the equation of the line.
<u>Slope</u>:
Let us convert the equation to slope - intercept form.


From the above equation, the slope is 
Since, the lines are perpendicular, the slope of the line can be determined using the formula,



Therefore, the slope of the equation is 
<u>Equation of the line:</u>
The equation of the line can be determined using the formula,

Substituting the point (6,9) and the slope
in the above formula, we get;

Simplifying the terms, we get;



Thus, the equation of the line is 
Answer:
(4x-7)(4x+7)
Step-by-step explanation:
This is called the Difference of Two Squares (DOTS).
These are the conditions that must be met for DOTS to work:
•There must be 2 terms.
•They must be separated by a negative sign.
•Each term must be a perfect square.
First, you have to find 2 square numbers that times together to make 16.
4×4=16
As we want 16x², you just need to make the 4 become 4x:
4x times 4x equals 16x²
Then, find 2 squares that times together to make 49.
7×7=49
You can put the two squares into a bracket with the x variable first, then the number:
(4x-7)(4x+7)
In the brackets, you always put the negative sign in the first one and then the plus sign in the second one.
You can double check your answer by expanding it back out again using FOIL:
F-First
O-Outer
I-Inner
L-Last
Times the First two in the two brackets:
4x times 4x equals 16x²
Times the Outer two:
4x times 7 equals 28x
Times the Inner two:
-7 times 4x equals -28x
Times the Last two:
-7 times 7 equals -49
Form an equation:
16x²+28x-28x-49
The 28x cancel out to leave:
16x²-49
Hope this helps :)
The area of the shaded region of this figure which can be calculated by the difference of the larger region from the smaller region would be 104 in.²
<h3>What is the area of the rectangle?</h3>
The area of the rectangle is the product of the length and width of a given rectangle.
The area of the rectangle = length × Width
Area of a larger rectangle
= Length × Width
= 18 × 12
= 216 in.²
Area of the smaller rectangle
Length = 12 - 2 - 2 = 8 in.
Width = 18 - 2 - 2 = 14 in.
= Length × Width
= 8 × 14
= 112 in.²
Area of the shaded region = Area of a larger rectangle - Area of the smaller rectangle
= 216 - 112
= 104 in.²
Hence, the area of the shaded region of this figure is 104 in.²
Learn more about the area;
brainly.com/question/2289492
The closest one is 5 to the power of 1 over 3.
sq rt of 3 = 1.732
sq rt of 5 = 2.236
5^1/3 = 1.667
5^1/6 = 0.833
5^2/3 = 8.333
5^3/2 = 62.5
Answer:
surface area of rectangle= 2[(l+w) +(l+h). +(w+h)]
hope it helps