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3 years ago

10 x 200 is how many tens

2 answers:

8 0

Since 10 times 200 is 2000 Therefore how many tens are in two thousand so you divide it and 2000 divided by10 will give you 200 So there are 200 tens in 10 times 200

melisa1 [442]3 years ago

7 0

10 x 200 is 2000
therefore there are 200 tens .

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What is the value of the square root of 36

irakobra [83]

6

Like how $6^{2}$ = 36

Square root does the opposite.

Edit: my mistake, it is 6

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3 years ago

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On the same winter morning, the temperature is -28f in Anchorage,Alaska and 65f in Miami,Florida. How many degrees warmer was it

Xelga [282]

C. 90 degrees warmer

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3 years ago

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M i crrect please explain

kvv77 [185]

I got -6/25 Sorry I can't really explain this try web2.0calc.com it helps alot with math.

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3 years ago

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How much more would $1,000 earn in 5 years in an account compounded continuously than an account compounded quarterly if the int

s2008m [1.1K]

Answer: There is a difference of $ 1.0228.

Explanation: Given, initial amount or principal = $ 1000,

Time= 5 years and given compound rate of interest = $3.7%

Now, Since the amount in compound continuously,

$A= Pe^{rt}$ , where, r is the rate of compound interest, P is the principal amount and t is the time.

Here, P=$ 1000, t=5 years and r= $3.7%,

Thus, amount in compound continuously , $A=1000e^{3.7\times5/100}$

⇒ $A=1000e^{18.5}=1000\times 1.20321844013=1203.21844013$

Therefore, interest in this compound continuously rate =1203.21844013-1000=203.21844013

now, Since the amount in compound quarterly,

$A=P(1+\frac{r/4}{100} )^{4t}$ , where, r is the rate of compound interest, P is the principal amount and t is the time.

Thus, amount in compound quarterly, $A=1000(1+\frac{3.7/4}{100} )^{4\times5}$

⇒ $A=1000(1+\frac{3.7}{400} )^{20}$

⇒ $A= 1202.19567617$

Therefore, interest in this compound quarterly rate=1202.19567617-1000=202.19567617

So, the difference in these interests=203.21844013-202.19567617=1.02276396 ≈1.0228

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3 years ago

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I'm studying a new bacteria that doubles in numbers every 25 minutes. If i start with 50 bacteria, how long until i have 5 milli

dimaraw [331]

Answer:

415.63 minutes

Step-by-step explanation:

Growth can be represented by the equation $A=A_0e^{rt}$ . We can find the rate at which it grows by using t=25 minutes and $\frac{A_{0}}{A} =2$ or double the amount at that time. The first step we always take is to divide $A_0$ by A.

$[tex]\frac{A_{0}}{A}=e^{rt}\\2=e^{r(25)}$

$2=e^{(25)r}$

To solve for r, we will take the natural log of both sides and use log rules to isolate r.

$ln 2=ln e^{(25)r}\\ln 2=25r (ln e)\\\frac{ln2}{25} =r$

We know $lne=1$ so we were able to cancel it out and divide both sides by 25.

We solve with a calculator $\frac{ln2}{25} =r\\0.0277=r$

We change 0.0277 into a percent by multiplying by 100 to get 2.77% as the rate.

The equation is $A=A_0e^{0.0277t}$ .

We repeat the step above substituting A=5,000,000, $A_0$ =50, and r=0.02777. Then solve for t.

$5000000=50e^{0.0277t}\\\frac{5000000}{50} =e^{0.0277t}\\100000=e^{0.0277t}\\ln100000=lne^{0.0277t}\\ln100000=0.02777t(lne)\\\frac{ln100000}{0.0277} =t$

t=415.63 minutes

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3 years ago