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2 years ago

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