Can someone please help me with this.

A science experiment begins with a bacterial population of 12. After 1 hour, the population is 18. After 2 hours, the population

Bess [88]

The first thing you should do is calculate how much is the percentage of growth per hour.
For this, we make the following rule of three:
12 ---> 100%
18 ---> x
We clear x:
x = (18/12) * (100)
x = 150%
Therefore, the population increases by 50% every hour.
We now look for the function that models this problem:
y = A (b) ^ x
Where:
y = total population
A = initial population
b = exchange rate
x: time (in hours, in this case)
Substituting values:
y = 12 * (1.5) ^ x
Let's check for 2 hours (x = 2):
y = 12 * (1.5) ^ 2
y = 27
Which is correct
Answer:
The function that models the problem is:
y = 12 * (1.5) ^ x

6 0

3 years ago

What is the domain of the function y=2√x-6

irga5000 [103]

The domain would be [6,infinity)

6 0

3 years ago

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X^2=7x+8<br><br><br><br>please help<br>

aivan3 [116]

Answer:x＝8

Step-by-step explanation:

8X8＝64 and 7X8+8 is 64, too.

6 0

3 years ago

If AC = 14, what is the measure of AD? Round your answer to the nearest hundredth.

nikklg [1K]

Part A

Answer: Choice D) $\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}$

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Explanation:

Let E be the intersection point of the diagonals. For any parallelogram, the diagonals bisect each other. This means that AE = EC = 2n+5 and it also means EB = DE = 4n-3

Focus on triangle AED. This triangle is a right triangle with angle AED as a 90 degree angle (any rhombus has its diagonals that are perpendicular to one another). The legs of this triangle are AE = 2n+5 and ED = 4n-3

Apply the pythagorean theorem to find hypotenuse AD

$(\text{leg1})^2+(\text{leg2})^2 = (\text{hypotenuse})^2\\\\(\text{AE})^2+(\text{ED})^2 = (\text{AD})^2\\\\(\text{AD})^2 = (\text{AE})^2+(\text{ED})^2\\\\\text{AD} = \sqrt{(\text{AE})^2+(\text{ED})^2}\\\\\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}\\\\$

This points to choice D as the final answer.

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Part B

<h3>Answer: 7.07</h3>

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Explanation:

If AC = 14, then half of this is AE = 7. The diagonals bisect each other, so they cut each other in half with E at the midpoint.

Since AE = 2n+5, we can solve for n like so

2n+5 = 7

2n = 7-5

2n = 2

n = 2/2

n = 1

Then we can find the length of AD

$\text{AD} = \sqrt{(2n+5)^2+(4n-3)^2}\\\\\text{AD} = \sqrt{(2*1+5)^2+(4*1-3)^2}\\\\\text{AD} = \sqrt{(2+5)^2+(4-3)^2}\\\\\text{AD} = \sqrt{7^2+1^2}\\\\\text{AD} = \sqrt{49+1}\\\\\text{AD} = \sqrt{50}\\\\\text{AD} \approx 7.0710678\\\\\text{AD} \approx 7.07\\\\$

Segment AD is roughly 7.07 units long. The other external sides of the rhombus (AB, BC and CD) are also this length because the four sides of any rhombus are the same length.

7 0

3 years ago

Find the slope of the line that passes through (10, 8) and (1, 4).

maria [59]

2/9 is the slope

Explanation:
8-4 = 4 over 10-1 = 9

5 0

3 years ago

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