Answer: 18
Step-by-step explanation:
Evaluate 7 r − 15 /s
Given that r = 3 and s = 5
7r - 15/s
Substitute 3 for r and 5 for s in the above expression
7(3) - 15/5
= 21 - 3
= 18
The complete question is
"Point P is the incenter of triangle ABC, PZ = 7 units, and PA = 12 units.
The radius of the incircle centered at point P is ? units."
The radius of the incircle centered at point P is 7 units.
We are given that point P is the incenter of the triangle, and is the center point of the incircle of the triangle.
<h3>What is incircle?</h3>
The incircle is defined as the largest circle that can be made in a triangle and is tangent to each side of the triangle.
Here, The radius of the circle is going to be a perpendicular line from point P to any side of the triangle.
In the triangle ,
PZ = PY = PX,
Each measure of value is equal to the radius of the circle.
Therefore, we already know that PZ = 7 units making the radius, r = 7 units.
Learn more about incircle;
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Although the number of new wildflowers is decreasing, the total number of flowers is increasing every year (assuming flowers aren't dying or otherwise being removed). Every year, 25% of the number of new flowers from the previous year are added.
The sigma notation would be:
∑ (from n=1 to ∞) 4800 * (1/4)ⁿ , where n is the year.
Remember that this notation should give us the sum of all new flowers from year 1 to infinite, and the values of new flowers for each year should match those given in the table for years 1, 2, and 3
This means the total number of flowers equals:
Year 1: 4800 * 1/4 = 1200 ]
+
Year 2: 4800 * (1/4)² = 300
+
Year 3: 4800 * (1/4)³ = 75
+
Year 4: 4800 * (1/4)⁴ = 18.75 = ~19 (we can't have a part of a flower)
+
Year 5: 4800 * (1/4)⁵ = 4.68 = ~ 5
+
Year 6: 4800 * (1/4)⁶ = 1.17 = ~1
And so on. As you can see, it in the years that follow the number of flowers added approaches zero. Thus, we can approximate the infinite sum of new flowers using just Years 1-6:
1200 + 300 + 75 + 19 + 5 + 1 = 1,600
At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.
So the amount of chlorine in the tank changes according to

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):


![\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%5Cdfrac%7Bc%28t%29%7D%7B%28200-3t%29%5E%7B5%2F3%7D%7D%5Cright%5D%3D0)


There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

![\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7Bc%28t%29%3D%5Cdfrac1%7B200%7D%5Csqrt%5B3%5D%7B%5Cdfrac%7B%28200-3t%29%5E5%7D5%7D%7D)