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11Alexandr11 [23.1K]
3 years ago
7

Is there a method that could prove that these two triangles are similar?

Mathematics
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

  A  Yes, a sequence of dilations and rigid motions carries one triangle onto the other

Step-by-step explanation:

Since two angles are marked with the same measures in each triangle, the triangles are similar by the AA postulate with no further action.

By some accounts it seems to be necessary to lay one figure on top of the other to demonstrate similarity. In order to do that, the smaller triangle would need to be dilated and translated (a rigid motion) to make it occupy the same space as the larger triangle.

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Answer:

square and rectangle

Step-by-step explanation:

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Axis of symmetry for f(x) = -2x2 + 20x -42
Ksenya-84 [330]
For the function f(x)=2x^2 +20x-42

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With the values substituted, we get...

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You are an assistant director of the alumni association at a local university. You attend a presentation given by the university
NNADVOKAT [17]

Answer:

(0.102, -0.062)

Step-by-step explanation:

sample size in 2018 = n1 = 216

sample size in 2017 = n2 = 200

number of people who went for another degree in 2018 = x1 = 54

number of people who went for another degree in 2017 = x2 = 46

p1 = x1/n1 = 0.25

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At 95% confidence level, z critical = 1.96

now we have to solve for the confidence interval =

<h2> p1 -p2 ± z*\sqrt{((1-p1)*p1)/n1 + ((1-p2)*p2/n2}</h2>

0.25 -0.23 ± 1.96*\sqrt{((1 - 0.25) * 0.25)/216 + ((1 - 0.23) *0.23/200}

= 0.02 ± 1.96 * 0.042

= 0.02 + 0.082 = <u>0.102</u>

= 0.02 - 0.082 = <u>-0.062</u>

<u>There is 95% confidence that there is a difference that lies between  - 0.062 and 0.102 on the proportion of students who continued their education in the years, 2017 and 2018.</u>

<u></u>

<u>There is no significant difference between the two.</u>

5 0
3 years ago
Can somebody help me please:(!
dlinn [17]
The greatest number is 28 people
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√(2x+1)<br> Evaluate the integral<br> 2<br> (2x + 1) In (2x + 1)<br> dx.
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Substitute y=\ln(2x+1) and dy=\frac2{2x+1}\,dx, so that

\displaystyle \int \frac2{(2x+1) \ln(2x+1)} \, dx = \int \frac{dy}y = \ln|y| + C = \boxed{\ln|\ln(2x+1)| + C}

6 0
1 year ago
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