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Morgarella [4.7K]
3 years ago
10

For any positive numbers a, b, and d, with b≠1, logb(a·d) = _____.

Mathematics
2 answers:
Katena32 [7]3 years ago
4 0

Answer:

logb a* logb d

        or

logb a - logb d

             or

logb(a*d)

9966 [12]3 years ago
3 0

Answer:

The value of log_b(a\cdot d) is log_b(a)+log_b(d).

Step-by-step explanation:

Let a, b, and d are positive numbers.

Then we need to find the value of log_b(a\cdot d).

According to the multiplication property of logarithm.log_m(xy)=log_mx+log_my

The given expression is

log_b(a\cdot d)

Using the multiplication property of logarithm, we get

log_b(a\cdot d)=log_b(a)+log_b(d)

Therefore the value of log_b(a\cdot d) is log_b(a)+log_b(d).

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3 years ago
A thermometer is taken from a room where the temperature is 20◦C to the
lord [1]

Answer:

a) T(2) = 8.265: at time t = 2 minutes the temperature will be 8.265 degress

b) 6 = T(3.55): the temperature will be 6 degrees at time t = 3.55 minutes.

Step-by-step explanation:

When dealing with temperature changes, it's best to work with Newton's Law of Cooling.

T(t) = T_s + Ce^{kt}

here:

T(t) : the temperature in the room.

T_s : ambient (or outdoor) temperature (that always remains constant, in our case: T_s = 5 )

C\,\text{and}\,k: are constants

Our conditions are provided:

1) T(0) = 20

2) T(1) = 12

using the first condition

T(0) = 5 + Ce^{k(0)}\\20 = 5 + C(1)\\C = 15

using the second condition:

T(1) = 5 + Ce^{k(1)}\\12 = 5 + Ce^{k}\\e^k = \dfrac{7}{C}

we can use our calculated value of C to find k

e^k = \dfrac{7}{15}\\k = \ln{(\dfrac{7}{15})}\\k = -0.7621

Finally we can put these constants back in the main equation:

T(t) = T_s + Ce^{kt}

T(t) = 5 + 15e^{-0.7621t} or T(t) = 5 + 15e^{\ln{(\frac{7}{15})t}

a) Reading after one more minute?

so it's asking:

T(2) = ?

T(2) = 5 + 15e^{\ln{(\frac{7}{15})(2)}}\\T(2) = \dfrac{124}{15} \approx 8.267

Hence, after one more minute the temperature of the room will be 8.267 degrees

b) When will it be 6 degrees?

T(t) = 6?

6 = 5 + 15e^{-0.7621t}\\\text{and solve for $t$}\\\\\dfrac{6-5}{15}=e^{\ln{(\frac{7}{15})}t}\\\ln{\left(\dfrac{1}{15}\right)} = \ln{\left(\dfrac{7}{15}\right)t}\\\ln{\left(\dfrac{1}{15}\right)} \div \ln{\left(\dfrac{7}{15}\right)} = t \approx 3.55\\

Hence at t = 3.55 minutes the temperature of the room will be 6 degrees.

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Look at the triangle ABC.
CaHeK987 [17]

Answer:

ahh pythagoras' theorem, my old nemesis...

Step-by-step explanation:

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2 years ago
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At this rate about how long would it take this competitor to complete a 30-mile race
dem82 [27]
Divide the distance by the average speed to find the time. 30 miles 20.04 mph 1.497 hrs. So it would take about 1.5 hours.
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Use distributive property to solve, show all steps <br> 3(2x-6)
zlopas [31]

Answer:

6x-18

Step-by-step explanation:

First we multiply everything in the parentheses by 3

3×2x=6x

3×(-6)=-18

And now we put it together-

6x-18

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