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IrinaVladis [17]
3 years ago
15

\[f(x)=\sqrt{x ^4-16x ^{2}}\]

Mathematics
1 answer:
marusya05 [52]3 years ago
4 0
f ( x ) =  \sqrt{ x^{4} -16 x^{2} }
a ) The domain:
x^4 - 16 x² ≥ 0
x² ( x² - 16 ) ≥ 0
x² - 16 ≥ 0
x² ≥ 16
x ∈ ( - ∞, - 4 ] ∪ [ 4 , + ∞ )
b ) f ` ( x ) = \frac{1}{2 \sqrt{x ^{4}-16 x^{2}  } } *  ( x^{4}-16 x^{2} )` =
= ( 2 x³ - 16 x ) / √(x^4 - 16 x²)
c ) The slope of the tangent line at x = 5:
f ` ( 5 ) = ( 2 * 125 - 16 * 5 ) / √ ( 625 - 400 ) = 170 / 15 = 34 / 3
The slope of the line normal to the graph at x = 5:
m = - 3 / 34 
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Bonjour, pouvez vous m'aidez svp ^^ Ci-contre, [MH] est une hauteur du triangle MAT. Éva affirme que le périmètre du triangle MA
Anna35 [415]

Cette question est incomplète.

Question complète

Exercer:

En face, [MH] est une hauteur du triangle MAT.

a.Calculer MH, puis HT.

b. Eva dit: "".

A-t-elle raison? Expliquer.

MA = 7,8cm

MT = 7,5cm

AH = 3cm

Answer:

La réponse d'Eva est incorrecte

Step-by-step explanation:

Nous résolvons les questions abive en utilisant le théorème de Pythagore

c² = a² + b²

a) Application de la formule ci-dessus:

Étape 1

Trouver MH

MA² = MH² + HA²

7,8² = MH² + 3²

MH² = 60,84-9

MH² = 51,84

MH = √51,84

MH = 7,2 cm

Étape 2

Résoudre pour HT

MT² = MH² + HT²

7,5² = 7,2² + HT²

HT² = 56,25 à 51,84

HT = √4,41

HT = 2,1 cm

b) La formule pour le périmètre du triangle MAT =

MA + MT + AH + HT

= 7,8 + 7,5 + 2,1 + 3 = 20,4 cm

Le périmètre du triangle MAT est de 20,4 cm

Donc, la réponse d'Eva est incorrecte

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