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blsea [12.9K]
4 years ago
10

Surface area = 3.14(rs + r2)

Mathematics
1 answer:
Oksanka [162]4 years ago
3 0
Part A:SA = 3.14(rs + r^2)SA = 3.14((2)(3) + (2)^2)
Part B:SA = 3.14((2)(3) + (2)^2)SA = 3.14(6 + 4)SA = 3.14(10)SA = 31.4SA = 31.4 units^2
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Anyone know what 5x-4=36 is? I’m looking to solve the (x).
k0ka [10]

Answer:

x=8

Step-by-step explanation:

add 4 to both sides

divide 40 by 5

then you are left with x=8

6 0
3 years ago
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Write the complex number in trigonometric form a+bi. 3(cos270+I sin 270)
Darya [45]
Simply get the value of the angle and distribute away.

\bf 3[cos(270^o)+i~sin(270^o)\implies 3[0+i( -1)]\implies 3-3i
4 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
If sintheta = cos theta then the value of acute angle is ​
Lelu [443]

Answer:

45 degrees

Step-by-step explanation:

Here, we want to get the value of the acute angle

From the question, we have it that the sin of the acute angle equals its cosine

Recall;

cos A = sin (90-A)

Hence, if A is 45, then Cos A and Sine A will be the same

This means the value of theta which is the acute angle is 45 degrees

3 0
3 years ago
Assume the question has a solution for z .. <br> solve for z<br> z = ____
saveliy_v [14]

Answer:

z=-(51p+84)/(d+p)

Step-by-step explanation:

-p*(51+z)=dz+84

-51p-pz=dz+84

-51p-84=dz+pz

z=-(51p+84)/(d+p)

6 0
3 years ago
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