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Naddika [18.5K]
2 years ago
11

PLEASE HELP WITH THIS

Mathematics
2 answers:
ArbitrLikvidat [17]2 years ago
8 0

Answer:

B

Step-by-step explanation:

The question can be written is

5^x = 75

Now take the log of both sides.

log(5x) = log(75)            Bring the x outside the brackets

x log(5) = log(75)           Divide by log 5

x = log(75)/log(5)

log 75 = 1.8751

log 5 = .8990

x = 1.8751/.8990 = 2.6826

marusya05 [52]2 years ago
7 0

Answer:

  (b)  2.68

Step-by-step explanation:

Your question has a link to a suitable calculator. All you need to do is <em>make use of it</em>. It tells you log₅(75) ≈ 2.68.

__

<em>Other ways to look at the problem</em>

If you recall that <em>a logarithm represents an exponent</em> of the base, you can ask yourself, what power 5 must be raised to in order to get 75. You know that 5^2 = 25, and 5^3 = 125, so the power will have to be between 2 and 3. There is only one answer choice in that range: 2.68.

__

The rules of logarithms let you rewrite the given expression as ...

  log₅(75) = log₅(3×5²) = log₅(3) +2·log₅(5) = log₅(3) +2

The log₅ of any number between 1 and 5 will be a positive fraction, so this sum must be between 2 and 3.

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Read 2 more answers
A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the se
stealth61 [152]

Answer:

L = 2*√2

w = √2

Step-by-step explanation:

Given:

A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the semicircle.

Find:

What are the dimensions of the rectangle with maximum​ area?

Solution:

- Let the length and width of the rectangle be L and w respectively.

- We know that Length L lie on the diameter base. So , L < 4 and the width w is less than 2 . w < 2.

- Using the Pythagorean Theorem, we relate the L with w using the radius r = 2 of the semicircle.

                           r^2 = (L/2)^2 + (w)^2

                           sqrt (4 - w^2 ) = L / 2

                           L = 2*sqrt (4 - w^2 )           L < 4 , w < 2

- The relation derived above is the constraint equation and the function is Area A which is function of both L and w as follows:

                          A ( L , w ) = L*w

- We substitute the constraint into our function A:

                          A ( w ) = 2*w*sqrt (4 - w^2 )

- Now we will find the critical points for width w for which A'(w) = 0

                         A'(w) = 2*sqrt (4 - w^2 ) - 2*w^2 / sqrt (4 - w^2 )  

                         0 = [2*sqrt (4 - w^2 )*sqrt (4 - w^2 )   - 2*w^2] / sqrt (4 - w^2 )  

                         0 = 2*(4 - w^2 )   - 2*w^2

                         0 = -4*w^2 + 8

                         8/4 = w^2

                         w = + sqrt ( 2 )   ..... 0 < w < 2

- From constraint equation we have:

                          L = 2*sqrt (4 - 2 )

                          L = 2*sqrt(2)

7 0
4 years ago
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