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3 years ago

7

If two sides of a triangle are 6 and 16, what is the range of the possible lengths of the third side? Enter your answer as doubl

e inequality. Answer: If we denote the third side as c, then
Please help

1 answer:

sveticcg [70]3 years ago

5 0

Answer:

10<c<22

Step-by-step explanation:

16-6=10

16+6=22

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Step-by-step explanation:

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2 years ago

What percent of 50 is 36?

mihalych1998 [28]

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2 years ago

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maksim [4K]

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6 1/6

Step-by-step explanation:

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2 years ago

How to make 2 6/11 into a improper fraction and plz show your work my brain is not that smart

maksim [4K]

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Step-by-step explanation:

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3 years ago

If x = a sin α, cos β, y = b sin α.sin β and z = c cos α then (x²/a²) + (y²/b²) + (z²/c²) = ?

Oduvanchick [21]

$\large\underline{\sf{Solution-}}$

<u>Given:</u>

$\rm \longmapsto x = a \sin \alpha \cos \beta$

$\rm \longmapsto y = b \sin \alpha \sin \beta$

$\rm \longmapsto z = c\cos \alpha$

Therefore:

$\rm \longmapsto \dfrac{x}{a} = \sin \alpha \cos \beta$

$\rm \longmapsto \dfrac{y}{b} = \sin \alpha \sin \beta$

$\rm \longmapsto \dfrac{z}{c} = \cos \alpha$

Now:

$\rm = \dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} }$

$\rm = { \sin}^{2} \alpha \cos^{2} \beta + { \sin}^{2} \alpha \sin^{2} \beta + { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha (\cos^{2} \beta + \sin^{2} \beta )+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha \cdot1+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha + { \cos}^{2} \alpha$

$\rm = 1$

<u>Therefore:</u>

$\rm \longmapsto\dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} } = 1$

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2 years ago