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miss Akunina [59]
3 years ago
12

A population of bacteria is growing at a rate of 20% per hour. If the population starts at 320, what is it an hour later?

Mathematics
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

384.

Step-by-step explanation:

The population is given by the expression;

Y(t) = Pr^{t}

Where;

Y(t) is the number of bacteria at time, t.

P is the starting amount.

r is the growth rate.

Given that, P = 320, r = 20% = 0.2, t = 1

Substituting the parameters into the equation;

Y(t) = Pr^{t}

Y(1) = 320 * 0.2¹

Y(1) = 320 * 0.2

Y(1) = 64.

Therefore, the population of bacteria increases hourly by 64.

An hour later; 320 + 64 = 384.

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Which function is non linear y-4x=1 y=2+6x^4 x-2y=7
ioda

Answer:

12

Step-by-step explanation:

43

8 0
3 years ago
If one bag of grass seed will cover 225 square feet, what is the minimum number of bags the haney family will need to cover thei
Veronika [31]

Answer:

9 bags

Step-by-step explanation:

First we will need to calculate the area of their entire backyard, which has a shape of trapezium.

So the formula for the area of a trapezium is A=\frac{(a+b)*h}{2}

a and b are the lenghts of the parallel sides and h is the height.

From the picture a=10*5=50\\b=9*5=45\\h=8*5=40

Then we plug into our formula

A=\frac{(a+b)*h}{2} =\frac{(50+45)*40}{2} = 1200 square feet

To get the number of bags required we just simply divide the aree of their entire yard by the space that 1 bag covers. We get:

\frac{1900}{225} =8.444... bags

This is more than 8 bags so they will need 9 bags to cover their entire yard.

8 0
3 years ago
-3|5x-6|+24&gt;=15<br> there is 2 solutions plz help me asap i need it in less than 2 hours
kodGreya [7K]

Answer:

5X+27=15

5x=12

x=5/12

I believe

4 0
3 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7
salantis [7]

Answer:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

Substitute in <em>x</em> = 0 once more:

\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0
3 years ago
One-third of a number t is less than or equal to 5
Kryger [21]

Answer:

t ≤ 15

Step-by-step explanation:

One third of 15 is equal to five (1/3 × 15 = 5)

One third of t is lesser than or equal to five (1/3 × 5 ≤ 5)

<em>So, t cannot be more than 15.</em><em> </em><em>It</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>1</em><em>5</em><em>,</em><em> </em><em>though</em><em>.</em>

∴t ≤ 15

7 0
3 years ago
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