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Morgarella [4.7K]
2 years ago
10

I don’t understand how this question

Mathematics
1 answer:
Vladimir79 [104]2 years ago
4 0

Answer:

<u>x = √51</u>

Step-by-step explanation:

The given figure is a <u>right triangle</u>.

<u>Using the Pythagorean Theorem to find x</u>

  • (Altitude 1)² + (Altitude 2)² = (Hypotenuse)²
  • (7)² + x² = (10)²
  • 49 + x² = 100
  • x² = 51
  • <u>x = √51</u>
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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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3 years ago
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UkoKoshka [18]
The answer is A
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7 0
4 years ago
rick deposited $6200 in a savings account with simple interest. three years later, the account held $7130 what was the interest
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1) Gathering the data

P=$6200

F=7310

t= 3 yrs

2) Since it is a simple interest, it is a linear application. So we can find it using this formula:

undefined

8 0
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