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mamaluj [8]
3 years ago
14

What are the new vertices of triangle QRS, shown, if the triangle is reflected across the line y = x, also shown? Question 6 opt

ions: A) Q′ = (–2,–3), R′ = (1,1), S′ = (2,–1) B) Q′ = (2,–4), R′ = (4,0), S′ = (3,–1) C) Q′ = (2,–5), R′ = (5,0), S′ = (2,–1) D) Q′ = (5,2), R′ = (0,5), S′ = (1,2)

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
6 0

Answer:

C

Step-by-step explanation:

Under a reflection in the line y = x

a point (x, y ) → (y, x ) , thus

Q(- 5, 2 ) → Q'(2, - 5 )

R(0, 5 ) → R'(5, 0 )

S(- 1, 2 ) → S'(2, - 1 )

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Digiron [165]

Answer:

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Step-by-step explanation:

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3 years ago
Mick is planning a party. He wants to serve pizza, chips, and soda. Mick discovers that he can purchase one pizza for $11.80, on
Temka [501]

Answer:

Mick will spend $191.55

Step-by-step explanation

$11.80 times 12 is 141.6- Pizza

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141.6+23.85+26.1=191.55

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What is 39 x (15÷3) - 16 using pemdas​
Pavel [41]

Answer:

179

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39*(15/3)-16

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195-16

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8 0
3 years ago
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I need help with this
evablogger [386]

Answer:

A) (-8, -16)

B) (0, 48)

C) (-4, 0), (-12, 0)

Step-by-step explanation:

A) the vertex is the minimum y value.

extremes of a function we get by using the first derivation and solving it for y' = 0.

y = x² + 16x + 48

y' = 2x + 16 = 0

2x = -16

x = -8

so, the vertex is at x=-8.

the y value is (-8)² + 16(-8) + 48 = 64 - 128 + 48 = -16

B) is totally simple. it is f(0) or x=0. so, y is 48.

C) is the solution of the equation for y = 0.

the solution for such a quadratic equation is

x = (-b ± sqrt(b² - 4ac)) / (2a)

in our case here

a=1

b=16

c=48

x = (-16 ± sqrt(16² - 4×48)) / 2 = (-16 ± sqrt(256-192)) / 2 =

= (-16 ± sqrt(64)) / 2 = (-16 ± 8) / 2 = (-8 ± 4)

x1 = -8 + 4 = -4

x2 = -8 - 4 = -12

so the x- intercepts are (-4, 0), (-12, 0)

6 0
3 years ago
What is the area of the triangle
ivanzaharov [21]

Answer:

24

Step-by-step explanation:

base (b) = 8

height (h) = 6

A = bh/2

= 8x6/2

= 48/2

= 24

4 0
3 years ago
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