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Mathematics

2 answers:

Nina [5.8K]3 years ago

8 0

Answer:

$y(x)=19*x$

Step-by-step explanation:

Given the height of <em>6 </em>DVD cases, you can know the height of a <em>single</em> DVD case, by dividing the total height, in the number of cases ( $\frac{114mm}{6} = 19mm$ for this problem).

Now, knowing the height of a <em>single</em> DVD case ( $19mm$ ), you can think the expression in this way: <em>If I already know the height of a single case, the height of any number of cases will be the result of multiplying the height of the case by the number of cases I want to stack.</em>

As the problem states, x is the number of cases in the stack, so, multiply the height of a single case for the number of cases (x).

This will result in the expression: $y(x)=19*x$

To know the height of a stack of 13 cases, just replace x=13 in the equation you just figured: $y(13)= 19* (13) = 247$ , but don't forget the unit is mm, so the final result is <u>247 </u><u><em>mm.</em></u>

Mama L [17]3 years ago

4 0

Y = 19x

The height of 13 would be 247mm

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Arlecino [84]

Answer:

D. Increasing both force and mass equally

Step-by-step explanation:

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3 years ago

Which expressions are equivalent to 2 ln a 2 ln b - ln a? Check all that apply. Ln ab2 - ln a ln a 2 ln b ln a2 ln b2 - ln a 2 l

marysya [2.9K]

Equivalent expressions are expressions with same simplified form. Equivalent expressions for the given expression are;

Expression 2: $\ln(a) + 2\ln(b) = \ln(a) + \ln(b^2) = \ln(ab^2)$
Expression 3: $\ln(a^2) + \ln(b^2) - \ln(a) = \ln(a^2b^2/a) = \ln(ab^2)$
Expression 5: $\ln(ab^2)$

<h3>What are equivalent expressions?</h3>

Those expressions who might look different but their simplified forms are same expressions are called equivalent expressions.

To derive equivalent expressions of some expression, we can either make it look more complex or simple. Usually, we simplify it.

<h3>What is logarithm and some of its useful properties?</h3>

When you raise a number with an exponent, there comes a result.

Lets say you get

$a^b = c$

Then, you can write 'b' in terms of 'a' and 'c' using logarithm as follows

$b = log_a(c)$

Some properties of logarithm are:

$log_a(b) = log_a(c) \implies b = c\\\\\log_a(b) + log_a(c) = log_a(b \times c)\\\\log_a(b) - log_a(c) = log_a(\frac{b}{c})\\\\log_a(b^c) = c \times log_a(b)\\\\log_b(b) = 1\\\\ log_a(b) + log_b(c) = log_a(c)$