Answer: 
Step-by-step explanation:
Using the values 
and 
, the slope is 
.
So, the equation is:
Answer:
the rate of change of the water depth when the water depth is 10 ft is; 
Step-by-step explanation:
Given that:
the inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.
We are meant to find the rate of change of the water depth when the water depth is 10 ft.
The diagrammatic expression below clearly interprets the question.
From the image below, assuming h = the depth of the tank at a time t and r = radius of the cone shaped at a time t
Then the similar triangles ΔOCD and ΔOAB is as follows:
( similar triangle property)
h = 2.5r
The volume of the water in the tank is represented by the equation:
The rate of change of the water depth is :
Since the water is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec
Then,
Therefore,
the rate of change of the water at depth h = 10 ft is:
Thus, the rate of change of the water depth when the water depth is 10 ft is; 
Answer:
Step-by-step explanation:
First you want to understand that every time a radio wave goes 300,000,000 meters a second passes, literally what meter per second means. of course it doesn't go that far, it goes 384,000 one way and then the same distance back, which makes 768,000. Under a third of the distance to make 1 second, so it will be less than one second.
You also want to understand the formula for speed. speed = distance / time. Whch also means, to find time you can change it to be time = distance / speed. And hey, we have distance and speed. 768,000 / 300,000,000 = .00256, I'll leave the significant figures to you unless you need help with them. but that's .00256 of a second. or .00256 s
Descending order...
5x^4 + 3x^3 + 6x^2 + 1...first term is 5x^4
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
