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Lina20 [59]
2 years ago
8

Determine if the two triangles are congruent.

Mathematics
1 answer:
goldenfox [79]2 years ago
6 0

Answer:

Yes

Step-by-step explanation:

Since they have 3 identical sides, they are congruent by SSS congruence.

HTH :)

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Please help me with this geometry question:((
Eddi Din [679]

Answer:

7.06

Step-by-step explanation:

Write a proportion using similar triangles.

DF / DG = DE / GE

x / 8 = 15 / 17

x = 7.06

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3 years ago
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Convert 140° from degree measure to radians
larisa86 [58]

Answer:

7pi/9

Step-by-step explanation:

140 degrees * (pi/180 degrees) = 7pi/9

3 0
2 years ago
Given the conversion factor 12 inches/ 1 foot, what is the line segment's length in feet?
serg [7]
103.2/12 is 8.6
A. 8.6 Feet
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2 years ago
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A bag contains red balls and white balls. If five balls are to be pulled from the bag, with replacement, the probability of gett
Ipatiy [6.2K]

Answer:

Percentage of balls which are red = 80%.

Step-by-step explanation:

Let the probability of drawing a red ball  be x then the probability of drawing a white ball is 1-x.

There are 5C3 = 10 ways of getting 3 reds in 5 draws so  the probability of this is 10* x^3 * (1 - x)^2.

The probability of getting 1 red in 5 draws = 5C1 * x (1-x)^4.

The  first probability is 32 times the last.

So we have the equation:

10x^3(1 - x)^2  /  5x(1 - x)^4  = 32

2x^2 / (1 - x)^2 = 32

2x^2 = 32(1 -x)^2

2x^2 = 32( 1 - 2x + x^2)

2x^2 = 32 - 64x + 32x^2

30x^2 - 64x  + 32 = 0

15x^2 - 32x + 16 = 0

(5x - 4)(3x - 4) = 0

x = 0.8, 1.333...

The probability must be < 1 so x = 0.8  = 80%.

6 0
3 years ago
How many permutations can be formed from all the letters in the word engineering?
laila [671]

Answer:

277,200

Step-by-step explanation:

To find the number of permutation we can form from the letters of the word "engineering", we first need to find the frequencies of the different letters present.

E = 3

G = 2

N= 3

I = 2

R = 1

Now that we have the frequencies, we count the number of letters in the word "engineering".

E N G I N E E R I N G

11 letters

Now we take the factorial of total number of letters and divide it by the number of repeats and their factorial

So we get:

\dfrac{11!}{3!2!3!2!1!}

We remove the 1! because it will just yield 1.

\dfrac{11!}{3!2!3!2!}

So the total number of permutations from the letters of the word "engineering" will be:

Total number of permutations = \dfrac{39,916,800}{144}

Total number of permutations = 277,200

3 0
3 years ago
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