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Harman [31]
3 years ago
7

List the sides of the triangle in order from shortest to longest.

Mathematics
2 answers:
olchik [2.2K]3 years ago
5 0

Answer:

MN < NO < OM

Step-by-step explanation:

∡ M = 180 -45-75=60°

45° opposite side - MN - shortest side

75° opposite side - OM - longest side

60° opposite side - NO

MN < NO < OM

katen-ka-za [31]3 years ago
3 0

Answer:

Angle M = 120 degrees

Shortest side = MN

Middle Side = MO

Longest side = ON

Step-by-step explanation:

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A right triangle has legs of lengths 6 and 15 inches. Find the length of the hypotenuse (nearest tenth).​
zubka84 [21]

Answer:

The length of the hypotenuse is 16.2 inches.

Step-by-step explanation:

c^{2} = a^{2} + b^{2}

c^{2} = 6^{2} + 15^{2}

c^{2} = 36 + 225

c^{2} = 261

\sqrt{c^{2} } = \sqrt{261}

c = 16.2

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3 years ago
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Encuentra el promedio,mediana,moda y rango de los siguientes datos: 14,16,21,25,16,12,18,17,19
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8 0
2 years ago
Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)
vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

x^{\frac{2}{3}}+y^{\frac{2}{3}}=4

We have to find the equation of tangent line to the given curve at point (-3\sqrt3,1)

By using implicit differentiation, differentiate w.r.t x

\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0

Using formula :\frac{dx^n}{dx}=nx^{n-1}

\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}

\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}

\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}

Substitute the value x=-3\sqrt3,y=1

Then, we get

\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}

\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}

Slope of tangent=m=\frac{1}{\sqrt3}

Equation of tangent line with slope m and passing through the point (x_1,y_1) is given by

y-y_1=m(x-x_1)

Substitute the values then we get

The equation of tangent line is given by

y-1=\frac{1}{\sqrt3}(x+3\sqrt3)

y-1=\frac{x}{\sqrt3}+3

y=\frac{x}{\sqrt3}+3+1

y=\frac{x}{\sqrt3}+4

This is required equation of tangent line to the given curve at given point.

8 0
3 years ago
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