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3 years ago

What is not preserved under dilation? Select all that apply.

Mathematics

1 answer:

marusya05 [52]3 years ago

3 0

Answer:

Hence, option D is correct (i.e. Distance is not preserved under dilation).

Step-by-step explanation:

<em>" A dilation is a transformation that produces an image that is the same shape as the original, but is a different size ".</em>

A) Angle measure:

Angle measures remains same as there is only a difference in their size the shape does not changes.

Hence angle measure is preserved.

B) Betweenness:

It is also preserved. since if any point in before dilation is between two points than after dilation it remain between them only.

C) Collinearity:

Points remain on the same line.

Hence, it is preserved.

D) Distance:

The distance is not preserved. since the length of the segment increase or decrease in dilation hence distance between two point also increase or decrease.

E) Proportionality:

In a dilation, the sides of the pre-image and the corresponding sides of the image are proportional. Hence it is also preserved.

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Find the length of the diagonal AD in the pentagon ABCDE shown in the coordinate plane.

Xelga [282]

Answer:

$\sqrt{73}$ units.

Therefore, Option C is correct.

Step-by-step explanation:

We have given a pentagon ABCDE

We need to calculate its length

That means we have to find the distance of AD

So, coordinates of A(1,6) and D(4,-2)

Using distance formula:

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here, $x_1=1,y_1=6,x_2=4,y_2=-2$

On substituting the values we will get:

$\sqrt{(4-1)^2+(-2-6)^2}$

On simplification:

$\sqrt{3^2+(-8)^2}$

$\sqrt{9+64}$

$\sqrt{73}$ .

Hence, the length of AD is $\sqrt{73}$ units.

Therefore, Option C is correct.

6 0

3 years ago

Give the exponential function A(x)=P(1+r)x, what value for r will make the function a growth function?

wolverine [178]

for this to be a growth function it has to be any number that is latger than zero

FOR EXAMPLE :-

50% which equals to 0.5

4 0

3 years ago

The Empirical Rule The following data represent the length of eruption for a random sample of eruptions at the Old Faithful geys

ad-work [718]

Answer:

(a) Sample Standard Deviation approximately to the nearest whole number = 6

(b) The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is.

(d) The actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) The percentage of eruptions that last less than 98 seconds using the empirical rule is 16%

(f) The actual percentage of eruptions that last less than 98 seconds is 15.866%

Step-by-step explanation:

(a) Determine the sample standard deviation length of eruption.

Express your answer rounded to the nearest whole number.

Step 1

We find the Mean.

Mean = Sum of Terms/Number of Terms

= 90+ 90+ 92+94+ 95+99+99+100+100, 101+ 101+ 101+101+ 102+102+ 102+103+103+ 103+103+103+ 104+ 104+104+105+105+105+ 106+106+107+108+108+108 + 109+ 109+ 110+ 110+110+110+ 110+ 111+ 113+ 116+120/44

= 4582/44

= 104.1363636

Step 2

Sample Standard deviation = √(x - Mean)²/n - 1

=√( 90 - 104.1363636)²+ (90-104.1363636)² + (92 -104.1363636)² ..........)/44 - 1

= √(199.836777 + 199.836777 + 147.2913224+ 102.7458678+ 83.47314049+ 26.3822314+ 26.3822314+ 17.10950413+17.10950413+ 9.836776857+ 9.836776857, 9.836776857+9.836776857+ 4.564049585+ 4.564049585+ 4.564049585+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 1.291322313+ 0.01859504133+ 0.01859504133+ 0.01859504133+ 0.7458677685+ 0.7458677685+ 0.7458677685+ 3.473140497+ 3.473140497+ 8.200413225+ 14.92768595+ 14.92768595+ 14.92768595+ 23.65495868+ 23.65495868+ 34.38223141+ 34.38223141+34.38223141+ 34.38223141+ 34.38223141+47.10950414+ 78.56404959+ 140.7458677+ 251.6549586) /43

= √1679.181818/43

= √39.05073996

= 6.249059126

Approximately to the nearest whole number:

Mean = 104

Standard deviation = 6

(b) On the basis of the histogram drawn in Section 3.1, Problem 28, comment on the appropriateness of using the Empirical Rule to make any general statements about the length of eruptions.

The use of Empirical Rule to make any general statements about the length of eruptions is empirical rules tell us about how normal a distribution and gives us an idea of what the final outcome about the length of eruptions is .

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

Mean = 104, Standard deviation = 6

For 68% μ - σ = 104 - 6 = 98, μ + σ = 104 + 6 = 110

For 95% μ – 2σ = 104 -2(6) = 104 - 12 = 92

μ + 2σ = 104 +2(6) = 104 + 12 = 116

Therefore, the percentage of eruptions that last between 92 and 116 seconds is 95%

(d) Determine the actual percentage of eruptions that last between 92 and 116 seconds, inclusive.

We solve for this using z score formula

The formula for calculating a z-score is is z = (x-μ)/σ

where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean = 104, Standard deviation = 6

For x = 92

z = 92 - 104/6

= -2

Probability value from Z-Table:

P(x = 92) = P(z = -2) = 0.02275

For x = 116

z = 92 - 116/6

= 2

Probability value from Z-Table:

P(x = 116) = P(z = 2) = 0.97725

The actual percentage of eruptions that last between 92 and 116 seconds

= P(x = 116) - P(x = 92)

= 0.97725 - 0.02275

= 0.9545

Converting to percentage = 0.9545 × 100

= 95.45%

Therefore, the actual percentage of eruptions that last between 92 and 116 seconds, inclusive is 95.45%

(e) Use the Empirical Rule to determine the percentage of eruptions that last less than 98 seconds

The empirical rule formula:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

For 68% μ - σ = 104 - 6 = 98,

Therefore, 68% of eruptions that last for 98 seconds.

For less than 98 seconds which is the Left hand side of the distribution, it is calculated as

= 100 - 68/2

= 32/2

= 16%

Therefore, the percentage of eruptions that last less than 98 seconds is 16%

(f) Determine the actual percentage of eruptions that last less than 98 seconds.

The formula for calculating a z-score is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For x = 98

Z score = x - μ/σ

= 98 - 104/6

= -1

Probability value from Z-Table:

P(x ≤ 98) = P(x < 98) = 0.15866

Converting to percentage =

0.15866 × 100

= 15.866%

Therefore, the actual percentage of eruptions that last less than 98 seconds is 15.866%

4 0

3 years ago

The family of functions y = ce^{-2x} + e^{-x} is solution of the equation y' + 2y = e^{-x}. Find the constant c which defines th

Rufina [12.5K]

Answer:

$c = 1190.2$

Step-by-step explanation:

We have the following solution:

$y(x) = ce^{-2x} + e^{-x}$

We want to find the value of x that satisfies the following condition:

y(3) = 3.

This means that when x = 3, y = 3. So

$y(x) = ce^{-2x} + e^{-x}$

$y(3) = ce^{-2*3} + e^{-3}$

$3 = ce^{-6} + e^{-3}$

$3 = ce^{-6} + 0.0498$

$ce^{-6} = 2.9502$

$c = \frac{2.9502}{e^{-6}}$

$c = 1190.2$

4 0

3 years ago

Can someone please help me?!!!

Afina-wow [57]

Answer:

sorry you out of luck man

3 0

2 years ago