ECH is an acute angle.
FCG is also an acute angle.
Area of a circle = πr^2
Area of a sector with angle x° = (x/360) πr^2
For the smaller sector,
x = 30°
r = 6 in
Required area of sector = (30/360) × π × 6^2 = 9.424 in^2
To the nearest hundreth, Area = 9.42 in^2
Answer:
h(x)= x^2+11x+30
Step-by-step explanation:
A quadratic function is in the form h(x) = ax^2 + bx + c.
Since the zeros are -6 and -5, take the opposite signs and add them to the variable x separately.
It should look like this: h(x)= (x+6)(x+5)
Since this is the factored form, we have to solve this equation further.
h(x)= (x+6)(x+5)
h(x)= x^2+6x+5x+30
h(x)= x^2+11x+30
10 inches of snow produce about an inch of water. However, this number can vary depending on the area, temperature, etc.
hope this helps
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:
(x-4)=0 ===> x=4,
so that now you have to plot 2 functions!
For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:
|x-4| = -(x-4)=4-x
Then:
for x<=4, y = -x+4-7 = -x-3
for x=>4, (x-4) is positive, so no changes:
y= x-4-7 = x-11,
Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11
In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:
x=4 and x= 3 for y=-x-3
And just x=5 for y=x-11
The reason is that the absolute value is continuous, so x=4 works for both:
x=4===> y=-4-3 = -7
x==4 ====> y = 4-11=-7!
abs() usually have a cusp int he point where it is =0
Hope it helps, despite being this long!
