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antiseptic1488 [7]
2 years ago
10

Between what pair of numbers is the product of 289 and 7?

Mathematics
1 answer:
marissa [1.9K]2 years ago
4 0
Because 289 is between 200 and 300, the product of 289 and 7 is between 200 x 7 and 300 x 7.

Therefore, the product of 289 and 7 is between 1,400 and 2,100.
Guest
1 year ago
thx
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Solve x2 = 12x – 15 by completing the square. Which is the solution set of the equation?
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For this case we have the following polynomial:
 x ^ 2 = 12x - 15

 The first thing to do is to place the variables on the same side of the equation.
 We have then:
 x ^ 2 - 12x = -15

 We complete the square by adding the term (b / 2) ^ 2 on both sides of the equation.
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 x ^ 2 - 12x + (-12/2) ^ 2 = -15 + (-12/2) ^ 2

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x ^ 2 - 12x + 36 = -15 + 36
 
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the solution set of the equation is:
 
x = 6 - \sqrt{21}
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The answer is 903!!!!!!
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There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them. 180 take calculus. T
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Answer:

150 students take physics.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents the students that take calculus.

-The set B represents the students that take physics

-The set C represents the students that take chemistry.

-The set D represents the students that do not take any of them.

We have that:

A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)

In which a is the number of students that take only calculus, A \cap B is the number of students that take both calculus and physics, A \cap C is the number of students that take both calculus and chemistry and A \cap B \cap C is the number of students that take calculus, physics and chemistry.

By the same logic, we have:

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

This diagram has the following subsets:

a,b,c,(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C), D

There are 360 people in my school. This means that:

a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360

The problem states that:

15 take calculus, physics, and chemistry, so:

A \cap B \cap C = 15

15 don't take any of them, so:

D = 15

75 take both calculus and chemistry, so:

A \cap C = 75

75 take both physics and chemistry, so:

B \cap C = 75

30 take both physics and calculus, so:

A \cap B = 30

Solution:

The problem states that 180 take calculus. So

a + (A \cap B) + (A \cap C) + (A \cap B \cap C) = 180

a + 30 + 75 + 15 = 180

a = 180 - 120

a = 60

Twice as many students take chemistry as take physics:

It means that: C = 2B

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

B = b + 75 + 30 + 15

B = b + 120

-------------------------------

C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)

C = c + 75 + 75 + 15

C = c + 165

----------------------------------

Our interest is the number of student that take physics. We have to find B. For this we need to find b. We can write c as a function o b, and then replacing it in the equations that sums all the subsets.

C = 2B

c + 165 = 2(b+120)

c = 2b + 240 - 165

c = 2b + 75

The equation that sums all the subsets is:

a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) + D = 360

60 + b + 2b + 75 + 30 + 75 + 15 + 15 = 360

3b + 270 = 360

3b = 90

b = \frac{90}{3}

b = 30

30 students take only physics.

The number of student that take physics is:

B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)

B = b + 75 + 30 + 15

B = 30 + 120

B = 150

150 students take physics.

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Answer:

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Well $16 times 16 would be $256.
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