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Blababa [14]
1 year ago
7

A biologist is studying the growth of a particular species of algae. She writes the following equation to show the radius of the

algae, f(d), in mm, after d days:
f(d) = 7(1.06)d

Part A: When the biologist concluded her study, the radius of the algae was approximately 13.29 mm. What is a reasonable domain to plot the growth function?

Part B: What does the y-intercept of the graph of the function f(d) represent?

Part C: What is the average rate of change of the function f(d) from d = 4 to d = 11, and what does it represent?
Mathematics
1 answer:
vitfil [10]1 year ago
4 0

Let's solve for d.

fd=(7)(1.06)d

Step 1: Add -7.42d to both sides.

df+−7.42d=7.42d+−7.42d

df−7.42d=0

Step 2: Factor out variable d.

d(f−7.42)=0

Step 3: Divide both sides by f-7.42.

d(f−7.42)f−7.42=0f−7.42

d=0f−7.42

Answer:

d=0f−7.42

PLEASE MARK ME AS BRAINLIEST

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Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k $

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean 210x^6y^4

We can deduce that this term will be located somewhere in the middle. So I will calculate k= 5; k=6 \text{ and } k =7.

For k=5

$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just x^{n-k} y^k

For k=6

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

5 0
2 years ago
A sphere has a diameter of 14 ft. Which equation finds the volume of the sphere?
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Answer:

The volume of the sphere is 1436 in³

The equation is    

V = ⁴⁄₃ * 3.14 * (7ft)³

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π = 3.14

V = ⁴⁄₃πr³

we replace with the known values

V = ⁴⁄₃ * 3.14 * (7ft)³

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V = 1436 in³

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Joshua started cycling at 5:15 pm By 8:09 pm, he has covered a distance of 7250 mWhat was Joshua's average speed during that tim
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The average speed of Joshua during that time is 2500 m/h.

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Hence, substituting the values we have,

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Dividing, we get,

speed=2500

Thus, the average speed of Joshua during that time is 2500 m/h.

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