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3 years ago

Divide x to the 5 eighths power divided by x to the 1 fourth power.

Mathematics

1 answer:

8_murik_8 [283]3 years ago

8 0

$\frac{x^{5/8}}{x^{1/4}} = x^{5/8-1/4}= x^{5/8-2/8}= x^{3/8} or \sqrt[8]{x^3}$

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Answer:

Step-by-step explanation:

w+13- $\frac{2w}{3}$ =16

w · $\frac{3}{3}$ - $\frac{2w}{3}$ +13=16

$\frac{w}{3}$ +13=16

$\frac{w}{3}$ =3

w=9

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2 years ago

Find the values of the sine, cosine, and tangent for ZA. (TOP OF TRIANGLE IS (A))

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$\bigstar\:{\underline{\sf{In\:right\:angled\:triangle\:ABC\::}}}\\\\$

AC = 7 m
BC = 4 m

⠀⠀⠀

$\bf{\dag}\:{\underline{\frak{By\:using\:Pythagoras\: Theorem,}}}\\\\$

$\star\:{\underline{\boxed{\frak{\purple{(Hypotenus)^2 = (Perpendicular)^2 + (Base)^2}}}}}\\\\\\ :\implies\sf (AB)^2 = (AC)^2 + (BC)^2\\\\\\ :\implies\sf (AB)^2 = (AB)^2 = (7)^2 = (4)^2\\\\\\ :\implies\sf (AB)^2 = 49 + 16\\\\\\ :\implies\sf (AB)^2 = 65\\\\\\ :\implies{\underline{\boxed{\pmb{\frak{AB = \sqrt{65}}}}}}\:\bigstar\\\\$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

☆ Now Let's find value of sin A, cos A and tan A,

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sin A = Perpendicular/Hypotenus = $\sf \dfrac{4}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{4 \sqrt{65}}{65}}$

⠀⠀⠀

cos A = Base/Hypotenus = $\sf \dfrac{7}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{7 \sqrt{65}}{65}}$

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tan A = Perpendicular/Base = ${\sf{\pink{\dfrac{4}{7}}}}$

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$\therefore\:{\underline{\sf{Hence,\: {\pmb{Option\:A)}}\:{\sf{is\:correct}}.}}}$

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