All categories

3 years ago

Quadrilateral ABCD is rotated 145° about point T. The result is quadrilateral A'B'C'D'. Which congruency statement is correct?

Mathematics

2 answers:

Vesna [10]3 years ago

6 0

The answer on e2020 is option letter A.

lilavasa [31]3 years ago

3 0

<span>The congruency statement should be that which shows the congruency of the corresponding angle and the sides. In the rotation of the polygon, the angle A is congruent with angle A' still. The same is true for all the angles and sides. The congruency statement is therefore, ABCD is congruent to A'B'C'D' The answer to this item is letter A. </span>

You might be interested in

traveled Downstream a distance of 33 Mi and then came right back. If the speed of the current was 12mph and the total trip took

spayn [35]

<span>Traveled Downstream a distance of 33 Mi and then came right back. If the speed of the current was 12 mph and the total trip took 3 hours and 40 minutes.
Let S = boat speed in still water then (s + 12) = downstream speed (s -12) = upstream speed
Given Time = 3 hours 40 minutes = 220 minutes = (220/60) h = (11/3) h Time = Distance/Speed
33/(s +12) + 33/(s-12) = 11/3 3{33(s-12) + 33(s +12)} = 11(s+12) (s -12) 99(s -12 + s + 12) = 11(</span> s^{2} + 12 s -12 s -144) 99(2 s) = 11(s^{2} -144) 198 s/11 = (s^{2} -144) 18 s = (s^{2} -144) (s^{2} - 18 s - 144) = 0 s^{2} - 24 s + 6 s -144 =0 s(s- 24) + 6(s -24) =0 (s -24) (s + 6) = 0 s -24 = 0, s + 6 =0 s = 24, s = -6 Answer) s = 24 mph is the average speed of the boat relative to the water.

7 0

3 years ago

What is a isometric dot paper

Nady [450]

Isometric graph paper is a triangular graph paper which uses a series of three guidelines forming a 60° grid of small triangles. The triangles are arranged in groups of six to make hexagons.

5 0

3 years ago

Read 2 more answers

A circle has a radius of 11cm find the area of a sector woth central angle that measures 225 degrees

LenKa [72]

Answer:

237.46 cm squared

Step-by-step explanation:

look at the image for explanation

3 0

3 years ago

The average score of all pro golfers for a particular course has a mean of 70 and a standard deviation of 3.0 (Think of these as

bija089 [108]

Answer:

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70

Standard Deviation, σ = 3

Sample size, n = 36

Let the average score of all pro golfers follow a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(score of the 36 golfers was between 70 and 71)

$\text{Standard error of sampling} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = \frac{1}{2}$

$P(70 \leq x \leq 71) = P(\displaystyle\frac{70 - 70}{\frac{3}{\sqrt{36}}} \leq z \leq \displaystyle\frac{71-70}{\frac{3}{\sqrt{36}}}) = P(0 \leq z \leq 2)\\\\= P(z \leq 2) - P(z \leq 0)\\= 0.977 - 0.500 = 0.477= 47.7\%$

$P(70 \leq x \leq 71) = 47.7\%$

0.477 is the probability that the average score of the 36 golfers was between 70 and 71.

8 0

3 years ago

Given: circle k(O), m AM=125°, m EF=31°, m∠MAF=75°. Find: m∠AME

Bess [88]

Answer:

58°

Step-by-step explanation:

If m AM=125°, then m∠AOM=125°.

If m∠MAF=75°, then m∠MOF=150° (because central angle MOF subtends on the same arc as inscribed angle MAF).

Thus,

m∠FOA=360°-150°-125°=85°.

If mEF=31°, then m∠EOF=31° (as central angle subtended on the arc EF).

Hence,

m∠EOA=m∠EOF+m∠FOA=31°+85°=116°.

Angle EOA is central angle subtended on arc EA, angle AME is inscribed angle subtended on arc AE, thus

m∠AME=1/2m∠EOA=58°.

7 0

3 years ago