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madam [21]
3 years ago
5

Pythagorean theorem!!! If c=7.4 and a=3.6 then is my answer b^2=sqrt 41.8 or just b=sqrt 41.8

Mathematics
1 answer:
Gre4nikov [31]3 years ago
4 0

Answer:

Step-by-step explanation:

a² + b² = c²

3.6² + b² = 7.4²

b² = 7.4² - 3.6² = 41.8

b = √41.8

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What are the potential solutions of log4x+log4(x+6)=2?
lions [1.4K]

The potential solutions of log_4x+log_4(x+6)=2 are 2 and -8.

<h3>Properties of Logarithms</h3>

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

  • Product Rule for Logarithms - log_{b}(a*c)=log_{b}a+log_{b}c
  • Quotient Rule for Logarithms - log_{b}(\frac{a}{c} )=log_{b}a-log_{b}c
  • Power Rule for Logarithms - log_{b}(a^c)=c*log_{b}a

The exercise asks the potential solutions for  log_4x+log_4(x+6)=2. In this expression you can apply the Product Rule for Logarithms.

                                  log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0

Now you should solve the quadratic equation.

 

 Δ=b^2-4ac=36-4*1*(-16)=36+64=100. Thus, x will be x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}. Then:

x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8

The potential solutions  are 2 and -8.

Read more about the properties of logarithms here:

brainly.com/question/14868849

4 0
2 years ago
A certain television is advertised as a 85 inch TV if the width of the TV is 67 how tall is the TV
Lunna [17]

Answer:

I believe the answer is 12

4 0
3 years ago
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Which list shows all the factors of 36?
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Answer:

1,2,3,4,6,9,12,18,36 Is correct one

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2 years ago
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What are the factors of 4
Nimfa-mama [501]
Factors are the numbers another number can be divided from. example-
8÷4=2, 4 is a factor of 8.

so the factors of 4 are 1,2and 4itself. like this-
4×1=4
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2×2=4
answer- 1,2 and 4.
7 0
3 years ago
There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student
Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0
3 years ago
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