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e-lub [12.9K]
2 years ago
6

Using the picture below, which angle is alternate exterior to angle 2?

Mathematics
1 answer:
kenny6666 [7]2 years ago
4 0

Answer:

alternate exterior means outside of the two parallel lines so your answer is 7 because alternate means other side of transversal

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Find the value c that completes the square for x^2-8x+c
Blababa [14]

Answer: 16

Step-by-step explanation:

x^2-8x+c

Let's complete the square.

c=(\frac{b}{2})^2

c=(\frac{-8}{2})^2

c=(-4)^2\\ c=16

7 0
3 years ago
In the given figure △ABC ≅△DEC. Which of the following relations can be proven using CPCTC ?
Serhud [2]

Option B:

\overline{A B}=\overline{D E}

Solution:

In the given figure \triangle A B C \cong \triangle D E C.

If two triangles are similar, then their corresponding sides and angles are equal.

By CPCTC, in \triangle A B C \ \text{and}\ \triangle D E C,

\overline{AB }=\overline{DE} – – – – (1)

\overline{B C}=\overline{EC} – – – – (2)

\overline{ CA}=\overline{CD} – – – – (3)

\angle ACB=\angle DCE  – – – – (4)

\angle ABC=\angle DEC  – – – – (5)

\angle BAC=\angle EDC  – – – – (6)

Option A: \overline{B C}=\overline{D C}

By CPCTC proved in equation (2) \overline{B C}=\overline{EC}.

Therefore \overline{B C}\neq \overline{D C}. Option A is false.

Option B: \overline{A B}=\overline{D E}

By CPCTC proved in equation (1) \overline{AB }=\overline{DE}.

Therefore Option B is true.

Option C: \angle A C B=\angle D E C

By CPCTC proved in equation (4) \angle ACB=\angle DCE.

Therefore \angle A C B\neq \angle D E C. Option C is false.

Option D: \angle A B C=\angle E D C

By CPCTC proved in equation (5) \angle ABC=\angle DEC.

Therefore \angle A B C\neq \angle E D C. Option D is false.

Hence Option B is the correct answer.

\Rightarrow\overline{A B}=\overline{D E}

5 0
3 years ago
A brick layer is hired to build three walls
hjlf
Answer is B because brick starts with a B
8 0
2 years ago
To prove that ΔAED ˜ ΔACB by SAS, Jose shows that StartFraction A E Over A C EndFraction = StartFraction A D Over A B EndFractio
Ira Lisetskai [31]

Answer:

A on Edg.

Step-by-step explanation:

Took the Test

5 0
3 years ago
Read 2 more answers
Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +
GenaCL600 [577]

<u>Answer: </u>

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

<u> Solution: </u>

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)  

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3)        ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2)        ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)    

On solving above equation for x    

(x+6) (x+3) =2(x+3) (2x+2)  

x+6 = 4x + 4    

x-4x = 4 – 6

x = \frac{2}{3}

Dimension of pool A

Length = x+6 = (\frac{2}{3}) +6 = 6.667m

Width = x +3 = (\frac{2}{3}) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2(\frac{2}{3}) + 6 = \frac{22}{3} = 7.333m

Width = 2x + 2 = 2(\frac{2}{3}) + 2 = \frac{10}{3} = 3.333m

Verifying the area:

Area of pool A = (\frac{20}{3}) x (\frac{11}{3}) = \frac{220}{9} = 24.44 meter

Area of pool B = (\frac{22}{3}) x (\frac{10}{3}) = \frac{220}{9} = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0
3 years ago
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