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dexar [7]
3 years ago
8

59 points - 12 points whats your grade

Mathematics
2 answers:
vazorg [7]3 years ago
7 0

Answer:

59-12 would be 47 points and by all means you would have an F

Step-by-step explanation:

59-12=47


lubasha [3.4K]3 years ago
6 0

Grade would be 47th......

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If log2 5 = k, determine an expression for log32 5 in terms of k.
lukranit [14]

Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

6 0
3 years ago
After John worked at a job for 10 years, his salary doubled. If he started at $x, his salary after 10 years is _____.
VARVARA [1.3K]
After ten years his salary is x2
6 0
3 years ago
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PLEASE ANSWER ALL OF THESE FOR 97 POINTS AND YOU ARE THE GOAT
harkovskaia [24]

Answer: It dosn't show anything for me. did you post it?

6 0
3 years ago
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There are three clubs at Redwood High School: Debate, Student council, and Key club. There are 1000 students in the school. And
hram777 [196]

Answer:

The number of students in the key club is 490

Step-by-step explanation:

The given parameters are;

The number of student in the school = 1000

The total number of student in the debate club = 310

The total number of student in the student council = 650

The total number of student who are in debate and student council = 170

The total number of student who are in both debate and the key club = 150

The total number of student who are in both student council and the key club = 180

The number of students who are in all three clubs = 50

Therefore, we have;

Let A represent the number of students in the debate club

Let B represent the number of students in the student council

Let C represent the number of students in the key club

We have;

n(A∪B∪C) = n(A) + n(B) + n(C) -  n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

Where;

n(A∪B∪C) = 1000

n(A) = 310

n(B) = 650

n(A∩B) = 170

n(B∩C) = 180

n(C∩A) = 150

n(A∩B∩C) = 50

Therefore;

n(C) = n(A∪B∪C) - (n(A) + n(B) -  n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C))

Substituting the values gives;

n(C) = 1000 - (310 + 650 -  170 - 180 - 150 + 50) = 490

The number of students in the key club, n(C) = 490.

5 0
3 years ago
PLEASE SHOW WORK....
Irina18 [472]

Answer:

Step-by-step explanation:

omg its says you cant have help

8 0
2 years ago
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