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MA_775_DIABLO [31]
3 years ago
13

HELPPP PLEASE

Mathematics
2 answers:
vazorg [7]3 years ago
7 0
You are very much welcomed
topjm [15]3 years ago
7 0
It should be 3 lmk if it’s wrong
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Rewrite 1/100,000,000 as a power of 10
yarga [219]
1/100,000,000 , note that 100,000,000 = 10⁸ , then:

1/100,000,000  = 1/10⁸
Remember that 1/aⁿ  = a⁻ⁿ

Hence 1/100,000,000 = 1/10⁸ = 10⁻⁸

3 0
3 years ago
Read 2 more answers
To find the product of 42.12 and 10³, move the decimal point in 42.12 __ places to the right because 10³ has __ zeros.
Yakvenalex [24]
<h3>To find the product of 42.12 and 10^3,  move the decimal point in 42.12 3 places to the right because 10^3 has 3 zeros</h3>

<em><u>Solution:</u></em>

Given that,

\text{ product of } 42.12 \text{ and } 10^3

Which means,

42.12 \times 10^3

Here, the exponent of 10 is positive ( which is 3)

When the exponent is positive, we have to move the decimal point to right

When you multiply a number by a power of 10, ( 10!, 10^2, and so on ) move the decimal point of the number to the right the same number of places as the number of zeros in the power of 10

Here, exponent is 3 , therefore move the decimal point right 3 places in 42.12

Therefore,

42.12 \times 10^3 = 42120

7 0
4 years ago
X+5=5<br> What is the value?
AveGali [126]
Answer: 1x or x

X=5-5
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4 0
3 years ago
Read 2 more answers
I need help on number 2.
evablogger [386]

<em>Greetings!</em>

This is how your double number line should look:

 (lbs)     0      1       2      3     4      5

              ↓-----↓-----↓-----↓-----↓-----↓

              0    3.5    7    10.5  14   17.5          (dollars)

Hope this helps!  

6 0
4 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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