All categories

2 years ago

Two parallel lines are cut by a transversal, as shown below. Solve for b. Then explain your reasoning.

Mathematics

1 answer:

Svetlanka [38]2 years ago

3 0

Answer:

b = 14

Step-by-step explanation:

Since both angles are diagonal each other, they are equal to each other.

5b = b + 56

-b -b (subtract b from both sides)

4b = 56

4 4 (divide both sides by 4 to solve for b)

b = 14

You might be interested in

Can someone help me on this problem

Marina CMI [18]

If a=b, the triangle ABC is an isosceles right triangle and the angle of interest is 45°.

sin(45°) = (√2)/2 . . . . . . selection C is appropriate

4 0

4 years ago

I am confusion. help me

timofeeve [1]

Multiple the two Collins

6 0

3 years ago

Find -2 2/5 + 4/5 need help please

diamong [38]

Step-by-step explanation:

-22/5 +4/5

The lcm of the denominators is 5

If you divide through u have - 22+4=-18

4 0

4 years ago

Match each function to its corresponding sequence.

givi [52]

The sequence and their functions are:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....
f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....
f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....

<h3>How to match the functions?</h3>

To do this, we simply set values for n and calculate the function values using the current value of n.

So, we have:

Function 1: f(n) = f(n - 1) + 6n - 5 where f(1) = 3

Let n = 2

f(2) = f(1) + 6(2) - 5 = 3 + 12 - 5 = 10

Let n = 3

f(3) = f(2) + 6(3) - 5 = 10 + 18 - 5 = 23

None of the sequence follows the pattern 3, 10, 23....

Function 2: f(n) = f(n - 1) + 2n - 1 where f(1) = 3

Let n = 2

f(2) = f(1) + 2(2) - 1 = 3 + 4 - 1 = 6

Let n = 3

f(3) = f(2) + 2(3) - 1 = 6 + 6 - 1 = 11

None of the sequence follows the pattern 3, 6, 11....

Function 3: f(n) = f(n - 1) + 6 where f(1) = 3

Let n = 2

f(2) = f(1) + 6 = 3 + 6 = 9

Let n = 3

f(3) = f(2) + 6 = 9 + 6 = 15

None of the sequence follows the pattern 3, 9, 15....

Function 4: f(n) = f(n - 1) + 7 where f(1) = 3

Let n = 2

f(2) = f(1) + 7 = 3 + 7 = 10

Let n = 3, 4 and 5

f(3) = f(2) + 7 = 10 + 7 = 17

f(4) = f(3) + 7 = 17 + 7 = 24

f(5) = f(4) + 7 = 24 + 7 = 31

So, we have:

f(n) = f(n - 1) + 7 ⇒ 3, 10, 17, 24, 31.....

Function 5: f(n) = f(n - 1) + 4n - 2 where f(1) = 3

Let n = 2

f(2) = f(1) + 4(2) - 2 = 3 + 8 - 2 = 9

Let n = 3, 4 and 5

f(3) = f(2) + 4(3) - 2 = 9 + 12 - 2 = 19

f(4) = f(3) + 4(4) - 2 = 19 + 16 - 2 = 33

f(5) = f(4) + 4(5) - 2 = 33 + 20 - 2 = 51

So, we have:

f(n) = f(n - 1) + 4n - 2 ⇒ 3, 9, 19, 33, 51.....

Function 6: f(n) = 2[f(n - 1)] where f(1) = 3

Let n = 2

f(2) = 2 * f(1) = 2 * 3 = 6

Let n = 3, 4 and 5

f(3) = 2 * f(2) = 2 * 6 = 12

f(4) = 2 * f(3) = 2 * 12 = 24

f(5) = 2 * f(4) = 2 * 24 = 48

So, we have:

f(n) = 2[f(n - 1)] ⇒ 3, 6, 12, 24, 48.....