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Y_Kistochka
2 months ago
15

Which of the following completes the proof? Triangles ABC and EDC are formed from segments BD and AC, in which point C is betwee

n points B and D and point E is between points A and C. Given: Segment AC is perpendicular to segment BD Prove: ΔACB ~ ΔECD Reflect ΔECD over segment AC. This establishes that ________. Then, ________. This establishes that ∠E'D'C' ≅ ∠ABC. Therefore, ΔACB ~ ΔECD by the AA similarity postulate. ∠ABC ≅ ∠E'D'C'; translate point E' to point A ∠ACB ≅ ∠E'C'D'; translate point E' to point B ∠ACB ≅ ∠E'C'D'; translate point D' to point B ∠ABC ≅ ∠E'D'C'; translate point D' to point A
Mathematics
2 answers:
zzz [12.3K]2 months ago
4 0

Response:

∠ACB ≅ ∠E'C'D'. move point E' to point A

Detailed explanation:

zzz [12.3K]2 months ago
3 0

Response: ∠ACB ≅ ∠E'C'D'; move point D' to point B

Detailed explanation:

This is simply my educated conjecture.

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Joan draws a bar chart to show the temperatures at midday on March 1st in five cities. Joan has made four mistakes in her diagra
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Answer:

Step-by-step explanation:

Characteristics of a bar graph include:

1). There must be uniform spacing between the bars or columns.

2). Each bar or column should have a consistent width.

3). All bars must share the same baseline.

4). The height of each bar corresponds to the data value.

Based on these criteria,

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8 0
3 months ago
Read 2 more answers
A box of 8 crayons costs $0.96. How much does
lawyer [12517]
0.96/8= $0.12 (per crayon)
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5 0
2 months ago
Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red
Zina [12379]
There are several possible outcomes. The initial composition of the urns is as follows: Urn 1 contains 2 red chips and 4 white chips, totaling 6 chips, whereas Urn 2 has 3 red and 1 white, amounting to 4 chips. When a chip is drawn from the first urn, the probabilities are as follows: for a red chip, it is probability is (2 red from 6 chips = 2/6 = 1/2); for a white chip, it is (4 white from 6 chips = 4/6 = 2/3). After the chip is transferred to the second urn, two scenarios can arise: if the chip drawn from the first urn is white, then Urn 2 will contain 3 red and 2 white chips, making a total of 5 chips, creating a 40% chance for drawing a white chip. Conversely, if a red chip is drawn first, Urn 2 will contain 4 red and 1 white chip, which results in a 20% chance of drawing a white chip. This scenario exemplifies a dependent event, as the outcome hinges on the type of chip drawn first from Urn 1. For the first scenario, the combined probability is (the probability of drawing a white chip from Urn 1) multiplied by (the probability of drawing a white chip from Urn 2), equaling 26.66%. For the second scenario, the probabilities yield a value of 6%.
8 0
2 months ago
Consider the area shown below. The height of the triangle is 8 and the length of its base is 3. We have used the notation Dh for
tester [12383]

Answer:

\text{Riemann sum }=\sum \frac{3}{8}(8-h)Dh

\text{Area =}\int_{a}^{b} \frac{3}{8}(8-h)Dh

Step-by-step explanation:

Given that the triangle's height measures 8 and the base length is 3, we can apply the concept of similar triangles to represent the base of the smaller triangle in relation to h.

The height of the smaller triangle will be (8-h).

Denote x as the base of the smaller triangle. Thus, by utilizing the properties of similar triangles, we can establish ratios of the corresponding sides as illustrated below:

\frac{8-h}{x} =\frac{8}{3} \\x=\frac{3}{8}(8-h)

This allows us to express the area of the small strip with length x and thickness Dh as follows:

DA=x*Dh\\DA=\frac{3}{8}(8-h)Dh

The desired Riemann sum can be articulated as:

\text{Riemann sum }=\sum \frac{3}{8}(8-h)Dh

The necessary areas can be represented as:

\text{Area =}\int_{a}^{b} \frac{3}{8}(8-h)Dh

Your remaining answers are accurate.:)

3 0
3 months ago
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