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2 years ago

Which division problem is equivalent to 38 38 ?

Mathematics

1 answer:

emmasim [6.3K]2 years ago

7 0

Just cut that, and do half of it.

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Find , , and if and terminates in quadrant .

ioda

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

STEP - BY - STEP EXPLANATION

What to find?

• sin2x

• cos2x

• tan2x

Given:

tanx = 2/3 = opposite / adjacent

We need to first make a sketch of the given problem.

Let h be the hypotenuse.

We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.

Using the Pythagoras theorem;

hypotenuse² = opposite² + adjacent²

h² = 2² + 3²

h² = 4 + 9

h² =13

Take the square root of both-side of the equation.

h =√13

This implies that hypotenuse = √13

We can now proceed to find the values of ainx and cosx.

Using the trigonometric ratio;

$\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}$ $\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}$

And we know that tanx =2/3

From the trigonometric identity;

sin 2x = 2sinxcosx

Substitute the value of sinx , cosx and then simplify.

$\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})$ $=\frac{12}{13}$

Hence, sin2x = 12/13

cos2x = cos²x - sin²x

Substitute the value of cosx, sinx and simplify.

$\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}$

Hence, cos2x = 5/13

tan2x = 2tanx / 1- tan²x

$\tan 2x=\frac{2\tan x}{1-\tan ^2x}$ $=\frac{2(\frac{2}{3})}{1-(\frac{2}{3})^2}$ $=\frac{\frac{4}{3}}{1-\frac{4}{9}}$ $=\frac{\frac{4}{3}}{\frac{9-4}{9}}$ $=\frac{\frac{4}{3}}{\frac{5}{9}}$ $=\frac{4}{3}\times\frac{9}{5}=\frac{4}{1}\times\frac{3}{5}=\frac{12}{5}$

$\tan 2x=\frac{\sin 2x}{\cos 2x}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$

Hence, tan2x = 12/5

Therefore,

sin2x =12/13

cos2x = 5/13

tan2x = 12/5

3 0

1 year ago

Solve.

Nikitich [7]

The answer is C: $10\text{ }^1/_2$ . Here are the details:

$\text{Equation:}\\ 6\text{ }^1/_3+10\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 6+10=16\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^1/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^2/_6+\text{ }^3/_6=\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{Add 'em up!}\\
16\text{ }^5/_6$

$\text{Equation:}\\
16\text{ }^5/_6+3\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{Start with the integers.}\\
16+3=19\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{...then with the fractions, but rewrite them first to make it easier.}\\
^5/_6+\text{ }^5/_6=\text{ }^{10}/_6\stackrel{\text{rewrite}}{\to}\text{ }^5/_3\stackrel{\text{rewrite}}{\to}1\text{ }^2/_3\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{Add 'em up!}\\
20\text{ }^2/_3$

$\text{Last equation:}\\ 20\text{ }^2/_3+5\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 20+5=25\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^2/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^4/_6+\text{ }^3/_6=\text{ }^7/_6\stackrel{\text{rewrite}}{\to}1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6\\
\\
\text{Add the integer and fraction together.}\\
25+1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6$

$26\text{ }^1/_6\stackrel{\checkmark}{=}26\text{ }^1/_6$

5 0

2 years ago

What is the missing justification in the proof?

nataly862011 [7]

The missing justification in the proof is
<span>B) Substitution property of equality

The expression for sin</span>² x and cos² x is substituted to the other side of the equation. Since sin x = a/c, then sin² x = a²/c². Similarly, since cos x = b/c, then cos² x = b²/c². Adding to two results to the third statement.

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3 years ago

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Predict the next three numbers in the pattern. 4 comma 2 comma 1 comma one half comma one fourth ,

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