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3 years ago

1. Are the following triangles similar? Justify your answer.

Mathematics

1 answer:

DerKrebs [107]3 years ago

5 0

Answer:

no they are not similar. this is because figure MCD had no corresponding sides to figure NLO. The two figures are not congruent or similar

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10) pls help i’ll give 20 points :)

NikAS [45]

Answer: The blank space in the equation is 4.

Step-by-step explanation: -6 x +2 is -12. -48/-12 is 4. Therefore, the blank space in the equation is 4.

3 0

3 years ago

You spin the spinner once. What is P(less than 3 or prime)? •3/5 •2/5 •1 •3/4

tatyana61 [14]

The answer is 3/4 I guess

7 0

4 years ago

How would I simplify and solve for X in this trigonometric equation? (Radians)

Phoenix [80]

Answer:

Please, refer to the images below

Step-by-step explanation:

We need to solve for x in the equation

cos (x+ pi) ^2 = sin (x)

cos (x+ pi) = - cos(x)

(-cos (x)) * (-cos (x)) = sin(x)

cos(x) ^2 = sin(x)

We know that

cos(x) ^2 + sin(x) ^2 = 1

cos(x) ^2 = 1 - sin(x) ^2

1 - sin(x) ^2 = sin(x)

sin(x) ^2 + sin (x) -1 = 0

Let A = sin(x)

A^2 + A - 1 = 0

(solutions attached in picture 1)

This means that

x = arcsin(A)

(solutions attached in picture 2)

8 0

3 years ago

If Isometry preserves distance and angle measurement ,which of the following are examples of Isometry?

3241004551 [841]

Answer:

C) Both reflection and translation

Step-by-step explanation:

An isometry of the plane is a linear transformation which preserves length. Isometrics includes rotation, translation , reflection ,glides and the identity map.

A geometry transformation is either rigid or non-rigid

The another name for a rigid transformation is 'isometry'.

The dilation is not isometry

7 0

4 years ago

Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with

skad [1K]

Parameterize S by the vector function

$\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

$\vec n = \dfrac{\partial\vec s}{\partial v} \times \dfrac{\partial \vec s}{\partial u} = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle$

The integral of the field over S is then

$\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv$

$\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv$

$\displaystyle = 128 \int_0^{\frac\pi2} \int_0^{\frac\pi2} \cos^2(u) \sin^3(v) \, du \, dv = \boxed{\frac{64\pi}3}$

8 0

2 years ago