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2 years ago

Given that lines a and b are parallel, what angles formed on line b when cut by the transversal are congruent with ∠2?

Mathematics

1 answer:

Kruka [31]2 years ago

3 0

Answer:

The angles formed on line b when cut by the transversal are congruent with ∠2 are $\angle{6}\text{ and }\angle{7}$

Step-by-step explanation:

Consider the provided information.

If transversal line crossed by two parallel lines, then, the corresponding angles and alternate angles are equal .

The angles on the same corners are called corresponding angle.

Alternate Angles: Angles that are in opposite positions relative to a transversal intersecting two lines.

∠2 and ∠6 are corresponding angles

Therefore, ∠2 = ∠6

∠2 and ∠7 are alternate exterior angles

Therefore, ∠2 = ∠7

Hence, the angles formed on line b when cut by the transversal are congruent with ∠2 are $\angle{6}\text{ and }\angle{7}$

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elena-14-01-66 [18.8K]

Answer:

Sample proportion ; phat

x / n

Step-by-step explanation:

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2 years ago

A dilation with a scale factor of 2 is applied to the three line segments shown. The resulting images are P′Q′¯¯¯¯¯¯¯, A′B′¯¯¯¯¯

ivanzaharov [21]

Answer:

Step-by-step explanation:

Formula to determine the scale factor,

Scale factor = $\frac{\text{Dimension of the image}}{\text{Dimension of the original}}$

2 = $\frac{P'Q'}{PQ}$

2 = $\frac{P'Q'}{2}$

P'Q' = 4 cm

Similarly, for the length of A'B',

2 = $\frac{A'B'}{AB}$

2 = $\frac{A'B'}{1.5}$

A'B' = 3 cm

For the length of M'N',

2 = $\frac{M'N'}{MN}$

2 = $\frac{M'N'}{3}$

M'N' = 6 cm

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2 years ago

PLZ ANSWER AND EXPLAIN IN DETAIL !!!!

xeze [42]

Answer:

720 liters

Step-by-step explanation:

In 2019 (a non-leap year)

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February has 28 days

March has 31 days

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2 years ago

Find the value of A that makes the following equation true for all values of x. (1/7)^x = A^-9x.

elena-14-01-66 [18.8K]

Answer:

Step-by-step explanation:

(1/7)^(x)=A^(-9x)

(1/7)^(x)=(A^(-9))^x

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A=(1/7)^9

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2 years ago

Solve the system of equations 2r + 2s = 50 and 2r – s = 17.

WARRIOR [948]

Eq1) 2r+2s=50
eq2) 2r-s=17

solve for s in equation2 (eq2)

-s=17-2r
s=-17+2r
Substitute s into equation1 (eq1)
2r+2(-17+2r)=50
2r-34+4r=50
6r-34=50
6r=50+34
6r=84
r=14
Substitute into either equation and solve for s
2(14)-s=17
28-s=17
-s=17-28
-s=-11
s=11

6 0

2 years ago