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luda_lava [24]
3 years ago
8

Which shows the correct way to use the rule ""multiply by the reciprocal"" for the expression below? 7 divided by StartFraction

5 Over 8 EndFraction StartFraction 1 Over 7 EndFraction times StartFraction 5 Over 8 EndFraction = StartFraction 5 Over 56 EndFraction StartFraction 7 Over 1 EndFraction times StartFraction 5 Over 8 EndFraction = StartFraction 35 Over 8 EndFraction StartFraction 1 Over 7 EndFraction times StartFraction 8 Over 5 EndFraction = StartFraction 8 Over 35 EndFraction StartFraction 7 Over 1 EndFraction times StartFraction 8 Over 5 EndFraction = StartFraction 56 Over 5 EndFraction
Mathematics
2 answers:
guajiro [1.7K]3 years ago
5 0

Given:

The expression is \dfrac{7}{\frac{5}{8}}.

To find:

The shows the correct way to use the rule ""multiply by the reciprocal"" for the given expression.

Solution:

According to the rule "multiply by the reciprocal", if a,b,c,d are four number, then

\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}

We have,

\dfrac{7}{\frac{5}{8}}

It can be written as

\dfrac{7}{\frac{5}{8}}=\dfrac{\dfrac{7}{1}}{\dfrac{5}{8}}

Using the rule "multiply by the reciprocal", we get

\dfrac{\dfrac{7}{1}}{\dfrac{5}{8}}=\dfrac{7}{1}\times \dfrac{8}{5}

\dfrac{\dfrac{7}{1}}{\dfrac{5}{8}}=\dfrac{7\times 8}{1\times 5}

\dfrac{\dfrac{7}{1}}{\dfrac{5}{8}}=\dfrac{56}{5}

Therefore, the correct option is D.

liq [111]3 years ago
5 0

Answer:

The correct answer is C.

Step-by-step explanation:

Hope this helps!

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Use these values in the pythagorean theorem to find side b

a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\

With respect to reference angle A, we have:

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\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\

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\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\

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\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

  • cosecant, abbreviated as csc, is the reciprocal of sine
  • secant, abbreviated as sec, is the reciprocal of cosine
  • cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{  ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

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Summary:

The missing side is b = \sqrt{33}

The 6 trig functions have these results

\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\

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