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2 years ago

Pls,pls,pls,help me I will give lots of pound 100+ points

Mathematics

1 answer:

Bond [772]2 years ago

4 0

Answer:

3log(10) -2log(5) ≈ 1.60206
no; rules of logs apply to any base. ln(x) ≈ 2.302585×log(x)
no; the given "property" is nonsense

Step-by-step explanation:

The given expression expression can be simplified to ...

3log(10) -2log(5) = log(10^3) -log(5^2) = log(1000) -log(25)

= log(1000/25) = log(40) . . . . ≠ log(5)

≈ 1.60206

Or, it can be evaluated directly:

= 3(1) -2(0.69897) = 3 -1.39794

= 1.60206

The properties of logarithms apply to logarithms of any base. Natural logs and common logs are related by the change of base formula ...

ln(x) = log(x)/log(e) ≈ 2.302585·log(x)

The given "property" is nonsense. There is no simplification for the product of logs of the same base. There is no expansion for the log of a sum. The formula for the log of a power does apply:

$\log(a)\log(b)=\log(a^{\log(b)})=\log(b^{\log(a)})$

Numerical evaluation of Mr. Kim's expression would prove him wrong.

log(3)log(4) = (0.47712)(0.60206) = 0.28726

log(7) = 0.84510

0.28726 ≠ 0.84510

log(3)log(4) ≠ log(7)

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sineoko [7]

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Step-by-step explanation:

Given

f(x)= 2^(x/2)

And

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We have to find (g+f)(x)

In order to find (g+f)(x), both the functions are added and simplified.

So,

(g+f)(x)= √(x-3)+2^(x/2)

The power x/2 can be written as a product of x*(1/2)

(g+f)(x)= √(x-3)+(2)^(1/2*x)

We also know that square root dissolves into power ½

(g+f)(x)=(x-3)^(1/2)+(2)^(1/2*x)

We can see that power ½ is common in both functions so taking it out

(g+f)(x)=(x-3+2^x)^(1/2)

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3 years ago

Denise has 6 hours to spend training for an upcoming race. She completes her training by running full speed the distance of the

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3 years ago

Describe how the graph is related to the graph of y = lxl.

Crazy boy [7]

Answer:

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Step-by-step explanation:

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3 years ago

A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the

Vedmedyk [2.9K]

(a) Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T - 75)$ where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get $\dfrac{dT}{T - 75} = k dt$ . Integrate both sides to get $\ln|T - 75| = kt + C$ .

Since $T(0) = 185$ , we solve for C:
$|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110$
So we get $\ln|T - 75| = kt + \ln 110$ . Use T(30) = 150 to solve for k:
$\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)$

So

$\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t} \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}$

But choose Positive because T > 75. Temp of turkey can't go under.

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)} = 136.929 \approx 137{}^{\circ}F$

(b)

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30} \\
100 = 75 + 110(15/22)^{t/30} \\
25 = 110(15/22)^{t/30} 
\frac{25}{110} = (15/22)^{t/30} \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}$

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Vitek1552 [10]

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