Y= -2x+4
The graph is linear
A vertical stretching is the stretching of the graph away from the x-axis.
A vertical compression is the squeezing of the graph towards the x-axis.
A compression is a stretch by a factor less than 1.
For the parent function y = f(x), the vertical stretching or compression of the function is a f(x).
If | a | < 1 (a fraction between 0 and 1), then the graph is compressed vertically by a factor of a units.
If | a | > 1, then the graph is stretched vertically by a factor of a units.
For values of a that are negative, then the vertical compression or vertical stretching of the graph is followed by a reflection across the x-axis.
Thus, the equation with the widest graph is 0.3x^2.
Answer: a = 64
Step-by-step explanation: the equation for area is a=lw. a = 16(4) which is equal to 64.
Step-by-step explanation:
the error:
it is stated that
subtract (5x)/(x+7) from both sides
multiply both sides by x + 7
7(x+7) = 49x
7x + 49 = 49x
subtract 7x from both sides to isolate x and its coefficient
49 = 42x
thus, this is only true when 49 = 42x. in order for these two equations to be equal, they must <em>always </em>be true, so this is wrong
the solution:
we want to express 7/x as (something) / (x+7). to do this, we can multiply 7/x by 1.
anything divided by itself = 1. thus, if we multiply both the numerator and the denominator by something that turns x into (x+7), we can do what we want to do.
(x+7)/x * x turns x into (x+7), so we multiply both the numerator and denominator by (x+7)/x to get
substitute this for 7/x in our original problem
Answer:
Required positive solution of the given quadratic equation is 9.
Step-by-step explanation:
Given Equation,
x² - 36 = 5x
We need to find positive solution of the given equation.
We solve the given quadratic equation using middle term split method.
x² - 36 = 5x
x² - 5x - 36 = 0
x² - 9x + 4x - 36 = 0
x( x - 9 ) + 4( x - 9 ) = 0
( x - 9 )( x + 4 ) = 0
x - 9 = 0 and x + 4 = 0
x = 9 and x = -4
Therefore, Required positive solution of the given quadratic equation is 9.
