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astra-53 [7]
3 years ago
15

Which expressions are equivalent to the one below?check all that apply.

Mathematics
2 answers:
Makovka662 [10]3 years ago
5 0

Answer:

A, B, and C


Step-by-step explanation:

The expression  log_{x}a=b means "x to the power b equal a"

<em>The expression log_5125+log_5125 means "5 raised to what power is 125" + "5 raised to what power is 125".</em>

we know 5 to the power 3 is 125, so the expression is 3 + 3 = 6


A.

log_{5}15,625 means 15 to "what power" gives us 15,625?

we know that 5 to the power 6 is 15625, so the expression log_{5}15,625 is equal to 6.

B.

The expression log_{5}(5^6) means 5 raised to what power is "5 raised to 5"? Simple, the answer is 6

C.

this is 6 (given)

D.

Whenever we don't have a base for the log, it is taken as "base 10". So the expression Log 15625 means "10 raised to what power is 15,625"? It is approximately 4.19


Thus we can see that Option A, B, & C all are equal to the expression given (6).

sineoko [7]3 years ago
3 0
The answers for the question shown above are the option A, the option B and the option C, which are:
 A.log5(15625)
<span> B.log5(5^6)
 C.6
 The explanation is shown below:
 By applying the logarithms properties, you have:
 A. </span><span>log5(125)+log5(125)=log5(125)(125)=log5(15625)
 B. </span>log5(125)+log5(125)=log5(15625)=log5(5^6)
 C. og5(125)+log5(125)=log5(15625)=log5(5^6)=6log5(5)=6
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3 years ago
Some of the students three scores is 231. If the first is 20 points more than the second, and the sum of the first two is 6 more
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Answer:

The first score is 109

Step-by-step explanation:

I am assuming that in the first sentence of the question, you meant:

Sum of the students three scores is 231...

First, let the scores of the first second and third student be a, b and c respectively. We are told that:

a + b + c = 231 . . . . . . . .(1)         (sum of students three scores is 231)

a = b + 20 . . . . . . . . . . . (2)        (the first is 20 points more than the second)

a + b = 6c . . . . . . . . . . . .(3)        (sum of the first two is 6 more times the third)

required, find a.

substituting the value of (a + b) in equation (3) into equation (1), we will have the following:

since a + b = 6c . . . (3)

a + b + c = 231 . . . . . (1), becomes,

(a + b) + c = 231

(6c)  + c = 231

7c = 231 (divide both sides by 7)

c = 231 ÷ 7 = 33

∴ c = 33

Next, from equation (2), we know that a = b + 20; this can also be written as:

a - 20 = b

∴ b = a - 20 . . . . . . . (4)

Finally, putting the value of b in equation (4) and the value of c calculated above into equation 1, ( a + b + c = 231), we have the following:

a + (a - 20) + 33 = 231

(a + a) - 20 + 33 = 231

2a + 13 = 231

2a = 231 - 13 = 218

a = 218 ÷ 2 = 109

∴ a = 109

we can also calculate for 'b' by substituting for the value of 'a' in equation 4

b = a - 20 = 109 - 20 = 89.

and to test if the values of a, b and c are correct:

a + b + c = 231

109 + 89 + 33 = 231

4 0
3 years ago
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