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lilavasa [31]
4 years ago
8

Danny draws a transversal, t, on two parallel lines AB and CD, as shown below:

Mathematics
2 answers:
DENIUS [597]4 years ago
8 0

Answer:

Corresponding angles formed by a transversal on parallel lines are congruent.

Genrish500 [490]4 years ago
7 0

Answer: second option.

Step-by-step explanation:

When two parallel lines are intersected by a third line (which is called "Transversal"), the angles that are of the same side of the transversal (but one located in interior and the other one located in the exterior), are known as "Corresponding angles".

Corresponding angles are congruent.

You can observe that the angle 2 and the angle 6 are Corresponding angles. Therefore, the missing justification is: "Corresponding angles formed by a transversal on parallel lines are congruent."

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1. Each row in a window display of floppy disk cartons contains two more boxes than the row above. The first row has one box. a.
Oduvanchick [21]
Each row has two more boxes than the row above. The first row has one box 

The boxes in a row form an arithmetic sequence with the first term, a₁ = 1 and the common difference, d=2.
The n-th term is
a_{n} = a+(n-1)d
The sum of n terms is
S_{n} = \frac{n}{2}(a_{1} + a_{n})

Answer:
The table will have the following:

Row Number:                        1   2   3    4    5    6
Boxes in the row:                  1   3   5    7    9    11
Total boxes in the display:    1   4   9  16  25  36
8 0
4 years ago
(9 – 23) + |47 – 16|
laiz [17]

You answer is: 17

Or otherwise B. is your correct answer.

(9 - 23) = -(14)

| 47 - 16| = 31

31 + (-14) = 17

5 0
3 years ago
A 6m ladder leans against a wall. The bottom of the ladder is 1.3m from the wall at time =0sec and slides away from the wall at
icang [17]

Answer:

- 0.100

Step-by-step explanation:

Length of the ladder,  H = 6 m

Distance at the bottom from the wall, B = 1.3 m

Let the distance of top of the ladder from the bottom at the wall is P

Thus,

from  Pythagoras theorem,

B² + P² = H²    .

or

B² + P² = 6²          ..............(1)      [Since length of the ladder remains constant]

at B = 1.3 m

1.3² + P² = 6²

or

P² = 36 - 1.69

or

P² = 34.31

or

P = 5.857

Now,

differentiating (1)

2B(\frac{dB}{dt})+2P(\frac{dP}{dt})=0

at t = 2 seconds

change in B = 0.3 × 2= 0.6 ft

Thus,

at 2 seconds

B = 1.3 + 0.6 = 1.9 m

therefore,

1.9² + P² = 6²

or

P = 5.69 m

on substituting the given values,

2(1.9)(0.3) + 2(5.69) × (\frac{dP}{dt}) = 0

or

1.14 + 11.38 × (\frac{dP}{dt}) = 0

or

11.38 × (\frac{dP}{dt}) = - 1.14

or

(\frac{dP}{dt}) = - 0.100

here, negative sign means that the velocity is in downward direction as upward is positive

5 0
4 years ago
2. What is the approximate value of b in the triangle below?
s344n2d4d5 [400]

Answer:

d) 17.6

Step-by-step explanation:

Use the law of Sines

SinA/a=SinB/b=SinC/c

6 0
4 years ago
Find the x-intercept of the parabola of with vertex (1,20) and the y-intercept (0,16). write your answer in this form: (x1,y1),(
svetoff [14.1K]
I assume that the parabola in this particular problem is one whose axis of symmetry is parallel to the y axis. The formula we're going to use in this case is (x-h)2=4p(y-k). We know variables h and k from the vertex (1,20) but p is not given. However, we can solve for p by substituting values x and y in the formula with the y-intercept:

(0-1)^2=4p(16-20)

Solving for p, p=-1/16.

Going back to the formula, we can finally solve for the x-intercepts. Simply fill in variables p, h and k then set y to zero:

(x-1)^2=4(-1/16)(0-20)
(x-1)^2=5
x-1=(+-)sqrt(5)
x=(+-)sqrt(5)+1

Here, we have two values of x

x=sqrt(5)+1 and
x=-sqrt(5)+1

thus, the answers are: (sqrt(5)+1,0) and (-sqrt(5)+1,0).
5 0
4 years ago
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