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fomenos
3 years ago
8

Geometry: CC 2015 > Chapter 2: Chapter 2 Test >

Mathematics
1 answer:
tankabanditka [31]3 years ago
7 0

Answer:

The area of ABCF is 2 and 2 × AB = 4

The statement proves that the conjecture is false

Step-by-step explanation:

A conjecture is a proposition or conclusion that is presumed to be true or correct but which is based on incomplete details or information

Given the length of the sides of rectangle ABDE, we have;

Length AB = 2 units

Length DE = 2 units

The area of ABDE = 2 × 2 = 4 unit²

Therefore, the area of the rectangle ABDE is equal to 2 × the length of either AB or DE

However, the are of rectangle ABCF = 2 × 1 = 2 unit²

While the area of rectangle 2 × the length of side AB = 2 × 2 = 4 unit², which is not equal to the number of square units in the area of the rectangle.

Therefore;

The area of ABCF is 2 and 2 × AB = 4

The statement (above) proves that the conjecture is false.

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If the point A (6,9) is translated 4 units left, then find the new coordinates of the point A.
Alex_Xolod [135]

Answer:

(2, 9 )

Step-by-step explanation:

A translation of 4 units left is equivalent to subtracting 4 from the value of the x- coordinate, that is

(6, 9 ) → (6 - 4, 9 ) → (2, 9 )

6 0
3 years ago
Can someone please tell me what angle x is
Inessa [10]

Answer:

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8 0
3 years ago
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Find -48a divided by 8 when a =1 . Write the answer in simplest form.<br><br> The solution is
love history [14]

Answer: -6

if a=1 then -48a=-48

so when we divide by 8 we ger -48/8 which is -6

7 0
3 years ago
For every 3 packages that Marcella wraps, she uses 12 of a roll of wrapping paper.
nikklg [1K]
For every 3 packages, she uses 1/2 a roll of wrapping paper

if she uses 2 1/2 rolls.....

(2 1/2) / (1/2) ....* 3
(5/2) / (1/2).....* 3
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7 0
3 years ago
I really need help quick
Natali [406]

Answer:

\sf x=2+\sqrt{6},\:x=2-\sqrt{6}

Explanation:

\sf y = x^2 -4x

given that y = 2

\sf x^2 -4x = 2

\sf x^2 -4x -2 = 0

using quadratic formula:

\sf x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}  \    \ \ when  \   ax^2 + bx + c = 0

\rightarrow \sf x_{1,\:2}=\dfrac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\cdot \left(-2\right)}}{2\cdot \:1}

\rightarrow \sf x_1=\dfrac{-\left(-4\right)+2\sqrt{6}}{2\cdot \:1},\:x_2=\dfrac{-\left(-4\right)-2\sqrt{6}}{2\cdot \:1}

\rightarrow \sf x=2+\sqrt{6},\:x=2-\sqrt{6}

in decimals:

\rightarrow \sf x=4.45 , \ -0.45

5 0
2 years ago
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