Circle with a line through it is square yards, a square with a line through it is square feet
No.
Yes, each line on a square is perpendicular to two other lines.
Draw one square, and draw one circle. Find the center in each, and draw one straight line vertically. Then find the middle once more but this time draw a line straight through horizontally. You have four equal parts in each shape.
solve the equation for y to get the slope.y=-2x-1/2substitute (3,3) into the equation/3=2(3)+band solve for b.-3=+by=2x-3 is the equation with the same slope(parallel) and goes through (3,3)
Circle with a line through it is square yards, a square with a line through it is square feet
An interior angle is an angle in the inside of two lines, and the two lines have a line through them. Look at the line through them. The opposite angles on each side of the line are equal. That should be enough info to solve your problem
No.
Very easy. Draw a square, draw a line through it then from above the line 4 times (draw it inside the square, not on the edges of the square) then you are done. it should be a 4x2 grid. 4x2 means 2 rows and 4 squares on each row.
Yes, each line on a square is perpendicular to two other lines.
The hard part is form a right angle. To do that draw a line with two points on it. Draw a circle around each point. Draw a line through the two points at which the circles intersect.
If y varies directly as x, then the graph of y against x must be a straight line through the origin.However, if y varies directly as the square of x, for example, then the graph of y against the square of x will be the straight line through the origin - not y against x.
i am not sure about how to solve the values for a theta graph but i know it looks like a spiral and you can plot these points if you draw a square and halve it and in each of the centres of the square is another point of theta.then you need the gradient of the line so if you have f(x)=θthen what is f(x)' which would be the gradient by the way f(x)' is another way of saying differentiated or dy/dθ
lets say you have a square of instance. You want to draw a line through the square and Have both sides of the square looking exactly the same the line. and that line u drew is ur axis of symmetry
Plug both points into the equation of a line, y =m*x + b and then solve the system of equations for m and b to get equation of the line through the points.
Any straight line through its centroid (centre of gravity).
You must solve the equations of the lines simultaneously. Represent each line in an equation of the form y=mx +b (where m represents the slope of the line, that is "rise over run"), then make substitutions using the info you have to solve for a pair of coordinates they share.