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rodikova [14]
3 years ago
6

If triangle CDE is dilated by a scale factor of 1/5 with a center of dilation at vertex E, what is the area of triangle C'D'E'?

Mathematics
2 answers:
timurjin [86]3 years ago
8 0
The correct answer would be Choice B: 4 square units.

When the scale factor is 1/5, that means the lengths of the sides are 1/5 of the original size. So instead of having a base of 20 and height of 10, the new triangle has a base of 4 and a height of 2.

The area of the triangle is: (4 x 2) / 2 = 4
Rainbow [258]3 years ago
3 0

Answer:

B) 4 units2

Step-by-step explanation:

The resultant triangle frmo the dilatation would have 1/5 of the size on each size, in the original triangle the base is 10 units and the height is 20 units, so 1/5 of 10 would be 2, and 1/5 of 20 would be 4, the resultant triangle from the dilatation at 1/5 would have an area of 4 squared units.

The area of the triangle is calculated with the next formula:

area=\frac{basexheight}{2} \\area=\frac{4x2}{2} \\Area=4u^{2}

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A shipping company claims that 95% of packages are delivered on time. A student wants to conduct a simulation
NISA [10]

The average number of packages the led to be selected in order to find a package that was not delivered on time is 2.2 option first is correct.

<h3>What is simple random sampling?</h3>

With simple random sampling, each component of the population has an equal chance of being selected for the sample.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

We have:

A shipping company claims that 95% of packages are delivered on time.

From the table:

00- 04 is the package not delivered on time.

05-99 - package delivered on time.

The first simulation's third number, 31, 64, 04, is where you can locate an item that wasn't delivered on time.

The second simulation's second number, 34, 02, is the location of a late-delivered delivery.

The third simulation is equal to 96,00 - the second number, which is the location of a late package.

The second number where to find a shipment not delivered on time in the fourth simulation is 31,04

Where to find a box that wasn't delivered on time in the fifth simulation: 98, 01, second number

The average number = (3 + 2 + 2 + 2 + 2)/5

= 11/5

= 2.2

Thus, the average number of packages the led to be selected in order to find a package that was not delivered on time is 2.2 option first is correct.

Learn more about the simple random sampling here:

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5 0
2 years ago
The sum of two numbers is 41. The larger number is one less than twice the smaller number. Find the numbers.
fgiga [73]

Answer:

14 and 17

Step-by-step explanation:

x+y=41

y=2x-1

x+(2x-1)=41

3x-1=41

3x=42

x=14

y=2(14)-1=27

5 0
3 years ago
Read 2 more answers
Whay will most likely happen to students test scores if the number of hours they use social media increases? A test score will i
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C as when they put more hours into the social media they will study less and less which leads to less test scores
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5. The temperature was
Vesna [10]

Answer:

6.2 is the difference

Step-by-step explanation:

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3 years ago
Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to f
tino4ka555 [31]

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :-  f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x)  to have a maximum or minimum at given point.

ii)  find first derivative f^{l} (x) and equating zero

iii) solve and find 'x' values

iv) Find second derivative f^{ll}(x) >0 then find the minimum value at x=a

v) Find second derivative f^{ll}(x) then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

<u>step1:</u>- find first derivative f^{l} (x) and equating zero

  f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)

f^{l}(x) = \frac{1}{x^2+1} (2x)  ……………(1)

f^{l}(x) = \frac{1}{x^2+1} (2x)=0

the point is x=0

<u>step2:-</u>

Again differentiating with respective to 'x', we get

f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}

on simplification , we get

f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}

put x= 0 we get f^{ll}(0) = \frac{2}{(1)^2}   > 0

f^{ll}(x) >0 then find the minimum value at x=0

<u>Final answer</u>:-

The minimum value of the given function is f(0) = 0

5 0
3 years ago
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