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2 years ago

E varies directly with the square root of C. If E=40 when C=25, find: C when E = 10.4

Mathematics

2 answers:

sukhopar [10]2 years ago

6 0

Answer: C = 1.69

Step-by-step explanation:

E is proportional to √C

To remove proportionality, introduce a constant (k).

E = k × √C

From question,

E = 40 and C = 25

So,

40 = k ×√25

40 = k × 5

k = 8

Now,

C = ?

E = 10.4

k = 8

E = k × √C

10.4 = 8 × √C

10.4 / 8 = √C

( 10.4 / 8 ) ^ 2 = C

C = 1.69

Ilia_Sergeevich [38]2 years ago

4 0

Answer:

Step-by-step explanation:

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shusha [124]

The value of b is -6.

Explanation:

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Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

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First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

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By the definition,

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Then

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Option B is false.

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$\sin C=\dfrac{AB}{AC},\\ \\\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

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$\dfrac{\cos A}{\tan C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{AB}{BC}}=\dfrac{BC}{AC}\neq \sin C.$

Option C is false.

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By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan C=\dfrac{AB}{BC}.$

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Then

$\dfrac{\cos(90^{\circ}-C)}{\tan A}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AB}}=\dfrac{AB^2}{AC\cdot BC}\neq \sin C.$

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