Jen and her friends need to consider the robot’s height and its width at its base. The robot must be short enough and narrow enough to navigate through several arches along the obstacle course. The director of the competition laid out components of the obstacle course on a coordinate plane. Instead of giving the width and maximum height of the arch, the director created an expression based on the location of the arch on the coordinate plane and gave each team this expression to represent the height of the arch, in inches, at any value of x along the arch: -x2 + 5x + 24. Draw a diagram of the archway modeled by the equation y = -x2 + 5x + 24. Find and label the y-intercept and the x-intercepts on the sketch. Then find and label the width of the archway at its base and the height of the archway at its highest point, assuming the base of the archway is along the x-axis. please help me any way you can.

I have the x and y int

sowwy i couldnt get threw this one

Step-by-step explanation:

x-Int (-8,0)

y-int (0,24)

hopes this helps

the answer is: the length is 5/6 yards.

give me a or a 5 star rating if this is correct!

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!