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Zolol [24]
3 years ago
13

Which side lengths form a right triangle?

Mathematics
1 answer:
STALIN [3.7K]3 years ago
3 0

Answer:

8, 15, 17  and \sqrt{2} , \sqrt{2}, 2

Step-by-step explanation:

c=\sqrt{a^2+b^2}

33=/=\sqrt{5^2+\sqrt{8}^2 }

17=\sqrt{8^2+15^2}

2=\sqrt{\sqrt{2}^2+\sqrt{2}^2  }

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What are the solution
AleksandrR [38]

Answer:

B

Step-by-step explanation:

We can get two equations from the inequality:

3x+2>9\\3x+2>-9

We just need to simplify both equations to get our answers:

3x+2>9\\3x>7\\x>\frac{7}{3}

3x+2>-9\\3x>-11\\x>\frac{-11}{3}

The two answers we get are:

x\frac{7}{3}

Which is also B.

3 0
2 years ago
Use natural logarithms to evaluate log7 2506
elixir [45]
Using the change in base property, we have \frac{ln2506}{ln7}=4.022 approximately.
6 0
4 years ago
10 + 9 • (−3) 2− (−1 + 8)
Vitek1552 [10]

Answer:

-51

Step-by-step explanation:

PEMDAS suggests you start with parenthesis

10+9•(-3)2-(7)

10-(27)2-(7)

10-54-7

-51

3 0
3 years ago
Read 2 more answers
Help me with this one please​
lesya [120]
For me, it’s easiest when i distribute the negative sign if i need to and then reorder to put the like terms together. and then solve.
(also sorry if it’s a little confusing with all the parentheses, i use them because it helps me organize everything)


21. (4x-9y) + (6x+10) + (8y-4)
= 4x + 6x - 9y + 8y + 10 - 4
= 10x - y + 6
—> D

22. (6x+9y-15) + (2x-9y+8)
= 6x + 2x + 9y - 9y - 15 + 8 (the 9y - 9y = 0, so you can leave it out of the final equation)
= 8x - 7
—> D

23. (9x^2-8x+3) - (5x^2-6x+4)
= 9x^2 - 8x + 3 - 5x^2 -(-6x) - 4
= 9x^2 - 5x^2 - 8x + 6x + 3 - 4 (remember that two - signs next to each other make a + sign)
= 4x^2 - 2x - 1
—> A

24. (9x^3-7x+8) - (5x^2+7x-10)
= 9x^3 - 7x + 8 - 5x^2 - 7x -(-10)
= 9x^3 - 5x^2 - 7x - 7x + 8 + 10
= 9x^3 - 5x^2 - 14x + 18
—> D

25. (6x+14y) - ((7x+5y) + (x-8y))
= (6x+14y) - (7x + x + 5y - 8y)
= (6x+14y) - (8x-3y)
= 6x + 14y - 8x -(-3y)
= 6x - 8x + 14y + 3y
= -2x + 17y
—> B
5 0
2 years ago
In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 g
KIM [24]

Answer:

3.52 \times 10^{-9} = 3.52 \times 10^{-7}\% probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

All girls from the first group:

20 students, so N = 20

10 girls, so k = 10

5 students will be selected, so n = 5

We want all of them to be girls, so we find P(X = 5).

P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163

All girls from the second group:

20 students, so N = 20

5 girls, so k = 5

5 students will be selected, so n = 5

We want all of them to be girls, so we find P(X = 5).

P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006

All girls from the third group:

20 students, so N = 20

8 girls, so k = 8

5 students will be selected, so n = 5

We want all of them to be girls, so we find P(X = 5).

P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}

3.52 \times 10^{-9} = 3.52 \times 10^{-7}\% probability that all the 15 students selected are girls

7 0
3 years ago
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