Answer:
(0, -9) (1, -5)
Step-by-step explanation:
We know from the equation y = 4x -9, that the slope of the line is 4, and the point that the line crosses the y-intercept is at -9. So, your first point would be at (0, -9) and if you go one value over on the x-coordinate, that means the y-coordinate will go 4 up. So your final point will be (1, -5). The point could also be (2, -1), (3, 3), (4, 7), (5, 11), etc.
I Hope That This Helps! :)
answer:
52
step-by-step explanation:
- we can use the 45-45-90 triangle theorem for this: x, x, x

this is because this is a square
- squares have two sides of the same length, and the diagonal is different
- the x's stand for the sides, whereas the x
stands for the diagonal
x
= 26
x = 13
- now multiply this by 4 because this is the length of one of the sides of the square. in order to find the perimeter we have to either add this four times or multiply by 4
13
X 4 = 52
In order to find the area of the area outside the circles and inside the square, you need the area of the square minus the area of all the circles.
I assume you know the area of the square (s^2= 4^2 = 16 square inches). But the area of the circles is tricky.
The area of a circle is pi times r squared

So you need the radius. Knowing the side length is four inches and that two circles span one side length (as in the picture), you can divide the side length by two to get the diameter of one circle. Then divide that by two again to get the radius. Ultimately the radius is one.
The area using the circle area equation is ultimately just pi. However, with four circles you need to multiply this by four to get four pi.
So, 16-4pi is approximately 3.44 inches squared is the answer.
Answer:
x^3 + 8y^6
Step-by-step explanation:
radius is the distance from the centre of a circle to any given point on its circumference, diameter is the length between two points that are the furthest possible from each other on the circle's circumference.
The diameter is twice the radius.
Therefore, if the radius is 12in then the diameter is 24in.