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3 years ago

In triangle MNO shown below, segment NP is an altitude:

Mathematics

1 answer:

Effectus [21]3 years ago

5 0

Answer: C. Definition of an Altitude

Step-by-step explanation:

Given: In triangle MNO shown below, segment NP is an altitude from the right angle.

Let ∠MNP=x

Then ∠PNO=90°-x

Therefore in triangle MNO,

∠MPN=∠NPO =90° [by definition of Altitude]

[Definition of altitude : A line which passes through a vertex of a triangle, and joins the opposite side forming right angles. ]

Now using angle sum property in ΔMNP

∠MNP+∠MPN+∠PMN=180°

⇒x+90°+∠PMN=180°

⇒∠PMN=180°-90°-x

⇒∠PMN=90°-x

Now, in ΔMNO and ΔPNO

∠PMN=∠PNO=90°-x

and ∠MPN=∠NPO =90° [by definition of altitude]

Therefore by AA similarity postulate, we have

ΔMNO ≈ ΔPNO

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3 years ago

Evaluate the triple integral ∭EzdV where E is the solid bounded by the cylinder y2+z2=81 and the planes x=0,y=9x and z=0 in the

dem82 [27]

Answer:

I = 91.125

Step-by-step explanation:

Given that:

$I = \int \int_E \int zdV$ where E is bounded by the cylinder $y^2 + z^2 = 81$ and the planes x = 0 , y = 9x and z = 0 in the first octant.

The initial activity to carry out is to determine the limits of the region

since curve z = 0 and $y^2 + z^2 = 81$

∴ $z^2 = 81 - y^2$

$z = \sqrt{81 - y^2}$

Thus, z lies between 0 to $\sqrt{81 - y^2}$

GIven curve x = 0 and y = 9x

$x =\dfrac{y}{9}$

As such,x lies between 0 to $\dfrac{y}{9}$

Given curve x = 0 , $x =\dfrac{y}{9}$ and z = 0, $y^2 + z^2 = 81$

y = 0 and

$y^2 = 81 \\ \\ y = \sqrt{81} \\ \\ y = 9$

∴ y lies between 0 and 9

Then $I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \int^{\sqrt{81-y^2}}_{z=0} \ zdzdxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{z^2}{2} \end {bmatrix} ^ {\sqrt {{81-y^2}}}_{0} \ dxdy$

$I = \int^9_{y=0} \int^{\dfrac{y}{9}}_{x=0} \begin {bmatrix} \dfrac{(\sqrt{81 -y^2})^2 }{2}-0 \end {bmatrix} \ dxdy$

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$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81x -xy^2} }{2} \end {bmatrix} ^{\dfrac{y}{9}}_{0} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81(\dfrac{y}{9}) -(\dfrac{y}{9})y^2} }{2}-0 \end {bmatrix} \ dy$

$I = \int^9_{y=0} \begin {bmatrix} \dfrac{{81 \ y -y^3} }{18} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \int^9_{y=0} \begin {bmatrix} {81 \ y -y^3} \end {bmatrix} \ dy$

$I = \dfrac{1}{18} \begin {bmatrix} {81 \ \dfrac{y^2}{2} - \dfrac{y^4}{4}} \end {bmatrix}^9_0$

$I = \dfrac{1}{18} \begin {bmatrix} {40.5 \ (9^2) - \dfrac{9^4}{4}} \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 3280.5 - 1640.25 \end {bmatrix}$

$I = \dfrac{1}{18} \begin {bmatrix} 1640.25 \end {bmatrix}$

I = 91.125

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3 years ago

Which mapping diagram does not represent a function? es A) 2 26 347 4 8 B) 2 37 4 18 C) B 2 6 3-47 4 8 D) 2 3 4 6 7 8

Jobisdone [24]

Answer:

I think it is A

Step-by-step explanation:

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3 years ago

What is the answer for m/1 =

Jobisdone [24]

<h3>Answer: 133</h3>

=============================================================

Explanation:

The quickest way to get this answer is to add the angles given to get 87+46 = 133

This is through the use of the remote interior angle theorem.

Note how the angles 87 and 46 are interior, or inside the triangle. And also, they are not adjacent to the exterior angle we want to find. So that's where the "remote" portion comes in.

----------------

The slightly longer method involves letting x be the measure of the missing interior angle of the triangle.

The three interior angles add to 180

87+46+x = 180

133+x = 180

x = 180 - 133

x = 47

The missing interior angle of the triangle is 47 degrees.

Angle 1 is adjacent and supplementary to this 47 degree angle, so,

(angle1)+(47) = 180

angle1 = 180-47

angle1 = 133 degrees

This example helps confirm that the remote interior angle theorem is correct.

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2 years ago