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Alenkasestr [34]

3 years ago

9

choose the most appropriate answer. temperature(10 seconds), temperature(10), and t(10) all represent which of the following? a.

the room temperature of 10 liquids b. the temperature of a liquid after it is heated for 10 seconds c. the 10th decimal place of a given temperature

1 answer:

Mice21 [21]3 years ago

7 0

B. the temperature of a liquid after it is heated for 10 seconds

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Please help please please help please

loris [4]

The correct answer should be D) 5/6

hope this helps :)

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3 years ago

The table below represents ordered pairs of a function. Which change could be made so that the relation in the table is no longe

kirill [66]

In order to be a function we need one x-value corresponding to one y-value which is shown in the table. if we can have the same x value corresponding to multiple y values then it is no longer a function it is a relation.

6 0

4 years ago

A garden table and bench cost $1020 combined. The garden table cost $70 more than the bench. What is the cost of the bench?

Dmitrij [34]

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8 0

4 years ago

Read 2 more answers

Rewrite the following expression using the properties of rational exponents. Be sure your answer is in simplest form.

Nookie1986 [14]

Answer:

$(2\cdot 6)^{\frac{3}{2}}=24(3)^{\frac{1}{2}}$

Step-by-step explanation:

The given expression is $(2*6)^{\frac{3}{2}}$

We multiply within the parenthesis to get:

$(12)^{\frac{3}{2}}$

Recall that: $a^{\frac{m}{n} } =\sqrt[n]{x^m}$

We apply this property to get:

$(12)^{\frac{3}{2}}=\sqrt{12^3}$

$(12)^{\frac{3}{2}}=\sqrt{12^2*12}$

This simplifies to

$(12)^{\frac{3}{2}}=12\sqrt{12}$

We can simplify further

$(12)^{\frac{3}{2}}=12\sqrt{4*3}=12*2\sqrt{3}=24\sqrt{3}$

We rewrite as rational exponent $24(3)^{\frac{1}{2}}$

3 0

3 years ago

Verify the identity. Show your work. cos(α - β) - cos(α + β) = 2 sin α sin β

s2008m [1.1K]

$\bf \textit{Sum and Difference Identities}
\\\\
cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)
\\\\
cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)\\\\
-------------------------------$

$\bf {cos(\alpha - \beta)-cos(\alpha+\beta)}=2sin(\alpha)sin(\beta)
\\\\\\
\stackrel{\textit{left-hand-side}}{[cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)]-[cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)]}
\\\\\\
\underline{cos(\alpha)cos(\beta)} + sin(\alpha)sin(\beta)-\underline{cos(\alpha)cos(\beta)} + sin(\alpha)sin(\beta)
\\\\\\
sin(\alpha)sin(\beta)+ sin(\alpha)sin(\beta)\implies 2 sin(\alpha)sin(\beta)$

7 0

3 years ago