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qaws [65]
3 years ago
6

Which polynomial is a perfect square trinomial?

Mathematics
2 answers:
yulyashka [42]3 years ago
4 0

Answer:

The third one

Step-by-step explanation:

The process would be figuring out the square root of the first and last term of each equation, after doing so, you would take that number (square root of first and last number) then you do 2(a)(b) which would be which should be the middle term. In this case 2(5)(4) = 40 which is correct.

zhenek [66]3 years ago
4 0

Step-by-step explanation:

my calculator got 36x*2-18-27/4

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Write the improper fraction as a mixed number in simplest form. 11/2
pantera1 [17]

Answer:

5 and 1/2

Step-by-step explanation: 2 goes into 11 5 times without passing it. That one that is left over would be the numerator and 2 as the demonator.

5 0
3 years ago
Through (5,1) , slope 4/5
Anettt [7]

Answer:

Step-by-step explanation:

3 0
3 years ago
Consider this equation: –3x – 5 2x = 6which is an equivalent equation after combining like terms?
Alla [95]
-3x-5+2x=6
2x-3x-5=6
-1x-5=6
-x-5=6
-x=11
x=11
an equivilent equaiton is x=11
7 0
3 years ago
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
Work out the size of angle x.<br> 38°<br> х<br> 101
KIM [24]

Answer:

41°

Step-by-step explanation:

the sum of the interior angles is 180 degree

so,

38 + x + 101 = 180

139 +x = 180

x = 180 - 139

x= 41°

5 0
3 years ago
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