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4 years ago

Name two ordered pairs that are located exactly six units from (2, -3).

2 answers:

6 0

Exactly six units from:

Def'n: Add or subtract 6 to either "x" or "y" in (x,y) = (2, -3)
P.S. (BTW, there are many more ways that are a lot harder to find two orders pairs six units from (2, -3). These are only the easiest.)

1. (2, -3 + 6) = (2, 3)
2. (2, -3 - 6) = (2, -9)
3. (2 + 6, -3) = (8, -3)
4. (2 - 6, -3) = (-4, -3)

Pick from these four. :3

Lapatulllka [165]4 years ago

5 0

(8,-3)
(2,3)

Hope it helps

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9 cos(sin¯¹(x)) = √81 – 81x²

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The equation 9cos(sin¯¹(x)) = √(81 – 81x²) is true since L.H.S = R.H.S

To answer the question, we need to know what an equation is

<h3>What is an equation?</h3>

An equation is a mathematical expression that show the relationship between two variables.

Given 9cos(sin¯¹(x)) = √(81 – 81x²), we need to show L.H.S = R.H.S

So, L.H.S = 9cos(sin¯¹(x))

= 9[√{1 - sin²(sin¯¹(x)}] (Since sin²y + cos²y = 1 ⇒ cosy = √[1 - sin²y])

9[√{1 - sin²(sin¯¹(x)}] = √9² × √{1 - sin²(sin¯¹(x)}]

= √[9² × {1 - sin²(sin¯¹(x)}]

= √[81 × {1 - sin²(sin¯¹(x)}]

= √[81 × {1 - x²}] (since sin²(sin¯¹(x) = [sin(sin¯¹(x)]² = x²)

= √(81 – 81x²)

= R.H.S

So, the equation 9cos(sin¯¹(x)) = √(81 – 81x²) is true since L.H.S = R.H.S

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2 years ago

Carlos biked StartFraction 139 over 8 EndFraction miles on Saturday and StartFraction 135 over 7 EndFraction miles on Sunday. On

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Step-by-step explanation:

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3 years ago

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at best buy a flat screen television with a regular price of 1,790 was reduced by $395 . Assuming 800 customers purchased the te

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3 years ago

Jamie needs to build a fence around his garden, as illustrated by polygon ABCDEF on the coordinate grid below. If each unit repr

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C. 26 yards

Step-by-step explanation:

Jamie's fence total length = perimeter of the polygon

Perimeter of the polygon = AB + BC + CD + DE + EF + FA

AB, FA and DE can be worked accordingly as shown below:

AB = |-5 - 0| = 5 units

FA = |5 - 2| = 3 units

DE = |1 -(-2)| = 3 units

BC, CD, and EF can be calculated using the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Distance between B(0, 5) and C(4, 2):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(0, 5) = (x_1, y_1)$

$C(4, 2) = (x_2, y_2)$

$BC = \sqrt{(4 - 0)^2 + (2 - 5)^2}$

$BC = \sqrt{(4)^2 + (-3)^2}$

$BC = \sqrt{16 + 9} = \sqrt{25}$

$BC = 5 units$

Distance between C(4, 2) and D(1, -2)

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(4, 2) = (x_1, y_1)$

$D(1, -2) = (x_2, y_2)$

$CD = \sqrt{(1 - 4)^2 + (-2 - 2)^2}$

$CD = \sqrt{(-3)^2 + (-4)^2}$

$CD = \sqrt{9 + 16} = \sqrt{25}$

$CD = 5 units$

Distance between E(-2, -2) and F(-5, 2):

$EF = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$E(-2, -2) = (x_1, y_1)$

$F(-5, 2) = (x_2, y_2)$

$EF = \sqrt{(-5 -(-2))^2 + (2 -(-2))^2}$

$EF = \sqrt{(-3)^2 + (4)^2}$

$EF = \sqrt{9 + 16} = \sqrt{25}$

$EF = 5 units$

Total length of the wall in yards = 5 + 5 + 5 + 3 + 5 + 3 = 26 yards

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