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3 years ago

Identify the percent amount and base in this problem.150 is 25 percent of what number.

Mathematics

1 answer:

NemiM [27]3 years ago

6 0

let's say that number is "x", so that is then the 100%.

if 150 is 25%, what is "x"?

$\bf \begin{array}{ccll} amount&\%\\ \cline{1-2} x&100\\ 150&25 \end{array}\implies \cfrac{x}{150}=\cfrac{100}{25}\implies \cfrac{x}{150}=4\implies x=600$

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stion 1 OT 5A researcher recorded the number of swans and the number of ducks in a lake every month. Function s represents the n

Anon25 [30]

Given functions are

$s(n)=2(1.1)^n+5_{}$ $d(n)=4(1.08)^n+3$

The total number of ducks and swans in the lake after n months can be determined by adding the functions s(n) and d(n).

$t(n)=s(n)+d(n)$

$t(n)=(2(1.1)^n^{}+5)+(4(1.08)^n+3)$

$t(n)=2(1.1)^n+5+4(1.08)^n+3$

$t(n)=2(1.1)^n+4(1.08)^n+5+3$

$t(n)=2(1.1)^n+4(1.08)^n+8$

Taking 2 as common, we get

$t(n)=2\lbrack(1.1)^n+2(1.08)^n+4\rbrack$

Hence The total number of ducks and swans in the lake after n months is

$t(n)=2\lbrack(1.1)^n+2(1.08)^n+4\rbrack$

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2 years ago

How to write four hundred twelve thousand, nine hundred eighty seven

Art [367]

The answer is 412,987

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3 years ago

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

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3 years ago

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shutvik [7]

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the notebook
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3 years ago

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astraxan [27]

Answer:

Step-by-step explanation:

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1 year ago