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cricket20 [7]
3 years ago
14

AB flies at 9 ft./s directly to A flower bed from its hive. The best days at the flower bed for 17 minutes, and then flies direc

tly back to the hive at 6 ft./s. It is away from the hive for a total of 20 minutes. A. What equation can you use to find the distance of the flowerbed from the hive. B. How far is the flowerbed from the hive
Mathematics
1 answer:
Kamila [148]3 years ago
7 0
Ask Socratic it helped
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Sliva [168]

Answer:

<h3>stay safe healthy and happy<u>.</u></h3>

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Twin​ brothers, Billy and​ Bobby, can mow their​ grandparent's lawn together in 53 minutes. Billy could mow the lawn by himself
podryga [215]
5353 minutes = 89.22 hours
Maybe the figures should be 53 and 20??

Let's let Billy = A
The formula to use for work problems is
Time = (A*B) / (A+B)

53 = ( (A +20)* A) / (A +20 + A) )
53 = ( A^2 + 20A ) / (2A + 20)
A^2 + 20A -106A -1060
A^2 -86A -1060
A = 96.935
Bobby Mows in 116.935 minutes

3 0
3 years ago
which of the following lengths of lines would be used to create a right triangle 6,9,12 5,12,13 4,8,16 7,15,19.7
marin [14]
If this is talking about the Pythagorean theorem then it would be 5,12,13 becuase because the equation for the triangle is
2 2 2
a + b = c

2 2 2
5 + 12. = 13
6 0
3 years ago
I’ve a always been a visual learner so can someone’s help me out with this online math
Maurinko [17]

Answer:

Question 2:

If the size of bacteria is 2000 initially and doubles after every half hour

It means in the 1st 30 min its size is 2000

And in 60 min its

30min=2000

60min=?

60×2000=120000

120000÷30

=4000

And in 9hrs

1hr=60min

9hrs=?

60×9=540min

30min=2000

540min=?

2000×540

=1080000÷30

=360000

7 0
3 years ago
Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species
kotegsom [21]

Answer:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

Step-by-step explanation:

The logistic equation is the following one:

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that P(0) = 80, K = 2000.

The number of fish tripled in the first year. This means that P(1) = 3P(0) = 3(80) = 240.

Using the equation for P(1), that is, P(t) when t = 1, we find the value of r.

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}

280*(2000 + 80(e^{r} - 1)) = 160000e^{r}

280*(2000 + 80e^{r} - 80) = 160000e^{r}

280*(1920 + 80e^{r}) = 160000e^{r}

537600 + 22400e^{r} = 160000e^{r}

137600e^{r} = 537600

e^{r} = \frac{537600}{137600}

e^{r} = 3.91

Applying ln to both sides.

\ln{e^{r}} = \ln{3.91}

r = 1.36

This means that the expression for the size of the population after t years is:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

4 0
3 years ago
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