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rewona [7]
3 years ago
13

Please help with 11,12,13

Mathematics
1 answer:
faltersainse [42]3 years ago
3 0

Problem 11

<h3>Answer:  h = 2A/b</h3>

------------------

Explanation:

We need to get h by itself. To do so, we first multiply both sides by 2. Then we divide both sides by b

A = (1/2)*b*h

2A = b*h

b*h = 2A

h = 2A/b

================================================

Problem 12

<h3>Answers:</h3>
  • Equation:   (n+2)/5 = 14
  • Solution to that equation:  n = 68

------------------

Explanation:

The number n is increased by 2 to get n+2

Then we divide by 5 to get (n+2)/5

This is set equal to 14 to get the equation (n+2)/5 = 14

Solving the equation would look like this

(n+2)/5 = 14

n+2 = 5*14 .... multiply both sides by 5

n+2 = 70

n = 70-2 .... subtract 2 from both sides

n = 68

================================================

Problem 13

<h3>Answer: Not a solution</h3>

------------------

Explanation:

We'll replace every copy of x with -3 and simplify

-2x + 5 > 13

-2*(-3) + 5 > 13

6 + 5 > 13

11 > 13

The last inequality is false because 11 is not greater than 13. Since the last inequality is false, this makes the first inequality false when x = -3.

Therefore, x = -3 is not a solution.

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Expand log_1/2(3x^2/2) using the properties and rules for logarithms.<br><br><br> Please help.
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Answer:

\frac{0.1761+2\log(x)}{-0.3010}

Step-by-step explanation:

Data provided:

\log_{1/2}(\frac{3x^2}{2})

now,

we know the properties of log functions as:

1) log(AB) = log(A) + log(B)

2) \log(\frac{A}{B}) = log(A) - log(B)

3) log(xⁿ) = n × log(x)

thus,

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3x^2) - \log_{1/2}(2)

or

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3)+\log_{\frac{1}{2}}(x^2) - \log_{1/2}(2)

or

using property 3

\log_{1/2}(\frac{3x^2}{2}) = \log_{1/2}(3)+2\log_{\frac{1}{2}}(x) - \log_{1/2}(2)

also,

\log_a(x)=\frac{\log(x)}{\log(a)}

thus,

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)}{\log(\frac{1}{2})}+2\times[\frac{\log(x)}{\log(\frac{1}{2}}]-\frac{\log(2)}{\log(\frac{1}{2})}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)+2\log(x)-\log(2)}{\log(\frac{1}{2})}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{\log(3)+2\log(x)-\log(2)}{\log(1)-log(2)}

now,

log(1) = 0

log(2) = 0.3010

log(3) = 0.4771

thus,

\log_{1/2}(\frac{3x^2}{2}) = \frac{0.4771+2\log(x)-0.3010}{0-0.3010}

or

\log_{1/2}(\frac{3x^2}{2}) = \frac{0.1761+2\log(x)}{-0.3010}

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Answer:

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