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Kipish [7]
3 years ago
8

Which statement correctly identifies the line of reflection?

Mathematics
2 answers:
Fofino [41]3 years ago
8 0

Answer:

D

Step-by-step explanation:

The black triangle is on -1, 1.

The red triangle is on 1, -1. The axles have flipped.

victus00 [196]3 years ago
5 0

Answer:

A. The triangles are reflected across the line y = -x

Step-by-step explanation:

Hope you have a great day :)

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x² + (y - 4)² = 25

Step-by-step explanation:

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x² + (y - 4)² = 25

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Let A=(-6,7), B=(-4,11), and C=(-2.10) be three points in the coordinate plane (A) Verify that the three points form a right tri
faust18 [17]

Answer:

  (A) see below

  (B) the right angle is at vertex B(-4,11)

  (1) 3x -4y = -46

  (E) the midpoint is (-4, 8.5)

Step-by-step explanation:

(A) The slope of line AB is Δy/Δx = (11-7)/(-4-(-6)) = 4/2 = 2. The slope of line BC is (10-11)/(-2-(-4)) = -1/2. The slopes of AB and BC have a product of 2(-1/2) = -1, so are the slopes of perpendicular lines. The points are distinct, and lines joining two pairs of them are at right angles, so the points form a right triangle.

(B) Point B(-4,11) is the point of intersection of perpendicular segments AB and BC, so is the location of the right angle.

(1) The slope of the hypotenuse AC is ...

  Δy/Δx = (10-7)/(-2-(-6)) = 3/4

In point-slope form, the equation for the line through point A with this slope is ...

  y -7 = 3/4(x +6) . . . . point-slope form of the equation of the hypotenuse

  4y -28 = 3x +18 . . . . multiply by 4

  -46 = 3x -4y . . . . .  . subtract 18+4y

  3x -4y = -46 . . . . . . . standard form equation of the hypotenuse

(E) The midpoint of the hypotenuse is the average of the endpoint coordinates:

  M = (A + C)/2 = (-6-2, 7+10)/2 = (-4, 8.5)

The midpoint of the hypotenuse is M(-4, 8.5).

8 0
4 years ago
How do you do these problems?
Angelina_Jolie [31]

Answer:

18e⁶

⁵/₁₂₈

Step-by-step explanation:

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) / (n+1)! (x − a)ⁿ⁺¹, and a < c < x.

f(x) = eˣ, a = 0, and n = 1.  Thus R₁ is:

R₁(x) = f"(z)/2! x²

R₁(x) = eᶻ/2 x²

|R₁| is a maximum when |f"(z)| is a maximum.  On the domain 0 < z < 6, that maximum is e⁶.  At x = 6, the upper bound of |R₁| is:

|R₁| = 18e⁶

This time, f(x) = 1 / √(1 + x) = (1 + x)^-½.  a = 0, and n = 2.

R₂(x) = f⁽³⁾(z)/3! x³

Find f⁽³⁾(x):

f'(x) = -½ (1 + x)^-³/₂

f"(x) = ¾ (1 + x)^-⁵/₂

f⁽³⁾(x) = -¹⁵/₈ (1 + x)^-⁷/₂

On the domain -½ < z < 0, |f⁽³⁾(z)| is a maximum at z = 0.

|f⁽³⁾(z)| = ¹⁵/₈

Therefore, at x = -½, the upper bound of R₂ is:

|R₂| = (¹⁵/₈)/6 |(-½)³|

|R₂| = ⁵/₁₂₈

8 0
3 years ago
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