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Oliga [24]
2 years ago
12

Given the figure below is a special type of trapezoid and WX || YZ, which angle pairs can be proven supplementary by the given i

nformation? Select all that apply.
∠W and ∠Z
∠W and ∠Y
∠X and ∠Y
∠X and ∠Z
∠W and ∠X

Mathematics
2 answers:
RSB [31]2 years ago
8 0

Answer:

The answer:

<W and <Z

<X and <Y

<W and <X


Step-by-step explanation:


svetlana [45]2 years ago
6 0

Answer:  The correct options are

(A) ∠W and ∠Z

(C) ∠X and ∠Y.

Step-by-step explanation:  Given that the figure is a special type of trapezoid and WX || YZ.

We are to select all the angle pairs that can be proven supplementary by the given information.

We know that

if two parallel lines are cut by a transversal, then the sum of the measures of interior angles on the same side of the transversal is 180°.

In the given trapezoid, we have

WX || YZ and WZ is a transversal, so ∠W and ∠Z are interior angles on the same side of the transversal WZ.

So,

m∠W + m∠Z = 180°.

This implies that ∠W and ∠Z are supplementary.

Similarly,

WX || YZ and XY is a transversal, so ∠X and ∠Y are interior angles on the same side of the transversal XY.

So,

m∠X + m∠Y = 180°.

This implies that ∠X and ∠Y are supplementary.

Therefore, the pairs of angles that can be proven supplementary with the given information are

∠W and ∠Z ;   ∠X and ∠Y.

Thus, (A) and (C) are correct options.

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5 0
2 years ago
Please help. I am getting in trouble so please try to be correct...
vichka [17]

Answer: The answer is A 17in2

Step-by-step explanation:

In the question it states that the triangles are congruent (both the same).

  first I found the area of the top orange triangle.

the formula to find the area of a triangle is \frac{1}{2} bh (base times Height).

so I did  \frac{1}{2} 3.23* 5.12  which gave me 8.27.

Since the triangles are congruent (the same) they would both have the same area along with base and height. so I multiplied 8.27 by 2 (because there are two triangles) and got 16.54 which rounds up to 17.

the question also stated to find the APPROXIMATE area (close to the actual, but not completely accurate or exact.)

7 0
2 years ago
The number of tickets purchased by an individual for Beckham College's holiday music festival is a uniformly distributed random
egoroff_w [7]

Answer:

The mean is 4.5 and the standard deviation is 1.44.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The mean of the uniform probability distribution is:

M = \frac{(a + b)}{2}

The standard deviation of the uniform probability distribution is:

S = \sqrt{\frac{(b-a)^{2}}{12}}

Uniformly distributed random variable ranging from 2 to 7.

This means that a = 2, b = 7.

So

M = \frac{(2 + 7)}{2} = 4.5

S = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(7 - 2)^{2}}{12}} = 1.44

The mean is 4.5 and the standard deviation is 1.44.

7 0
1 year ago
The random variable X is normally distributed with mean 82 and standard deviation 7.4. Find the value of q such that P(82 − q &l
mezya [45]
P(82 - q < x < 82 + q) = 0.44
P(x < 82 + q) - P(82 - q) = 0.44
P(z < (82 + q - 82)/7.4 - P(z < (82 - q - 82)/7.4) = 0.44
P(z < q/7.4) - P(z < -q/7.4) = 0.44
P(z < q/7.4) - (1 - P(z < q/7.4) = 0.44
P(z < q/7.4) - 1 + P(z < q/7.4) = 0.44
2P(z < q/7.4) - 1 = 0.44
2P(z < q/7.4) = 1.44
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8 0
1 year ago
Triangle ABE is similar to triangle ACD. AED and ABC are straight lines. EB and DC are parallel. AE = 5 cm, BC = 4.5 cm, BE = 4
lianna [129]

Answer:

y = 16x/65

Step-by-step explanation:

Given:

Triangle ABE is similar to triangle ACD. AED and ABC are straight lines

EB and DC are parallel

The area of quadrilateral BCDE = xcm²

The area of triangle ABE = ycm²

Find attached the diagram from the above information.

In similar triangles, the ratio of their corresponding angles are equal.

Also, the ratio of the area of the two triangles = square of ratio of the corresponding sides of the two triangles.

Area ∆ACD/area of ∆ABE = (DC/EB)²

Area ∆ACD/area of ∆ABE = [(area of quadrilateral BCDE +

area of ∆ABE)]/(area of ∆ABE)

(x+y)/y = (DC/EB)²

(x+y)/y = (9/4)²

x+y = (81/16)y

x = (81/16)y - y

x = (81y - 16y)/16

x = 65y/16

Making y subject of formula

16x = 65y

y = 16x/65

An expression for y in terms of x:

y = 16x/65

3 0
1 year ago
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