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Orlov [11]
3 years ago
12

What is the solution of a - 1 > 11

Mathematics
1 answer:
ozzi3 years ago
8 0
A>12!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Which is greater, 1/4 or 9/20
Zina [86]
9/20

You need to have the same denominator so multiply it by 5 on the bottom and top to get 5/20 then you can compare it 5/20 < 9/20
6 0
3 years ago
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Find the area of a parallelogram with sides of 6 and 12 and an angle of 60°.
ASHA 777 [7]
I think I got it. Correct me if I am wrong.

Parallelogram diagram I believe down below. We must find the height and then the area using Pythagorean theorem. Since the green shaded part is a 30-60-90 triangle, the base is 1/2 the hypotenuse, therefore it is 3. Now we calculate the height with it.

A^2 + B^2 = C^2
A^2 + 3^2 = 6^2
    A^2 + 9 = 36
          A^2 = 27
              A = 3√3

Therefore the height is 3√3

Now calculate the area using A = bh

A = bh
   = (12)(3√3)
   = 36√3

So the area is 36√3 square units.

I cannot be sure of this answer because you did not provide a diagram.

4 0
3 years ago
Simplify the expression (⎯+.)+(+.).
Korolek [52]
(-)(+)+(+)
=(-)(+)
=(-)
4 0
3 years ago
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The sum of four consecutive integers is 390. Find the integers (Please show your work)
AysviL [449]

Step-by-step explanation:

The sum of three consecutive even integers is 390. What are the integers?

 

   In order to ensure the numbers are even I generally call the first

   integer 2n  because 2 multiplied by any number will always be even.

   Our consecutive even integers are:  2n, 2n+2, 2n+4

 

   2n + 2n+2 + 2n+4 = 390

   6n + 6 = 390

   6n = 384

   n = 64

 

  Our 1st integer is 2n = 2(64) = 128

  Our integers are 128, 130, 132

 

Check:  128+130+132 = 390 which is correct.

 

2) The sum of three consecutive odd integers is 99. What are the integers?

 

    Same logic applies:  An odd integer will always result from 2n+1

    So we will call our 3 consecutive odd integers   2n+1, 2n+3, 2n+5

 

   2n+1 + 2n+3 + 2n+5 = 99

   6n+9 = 99

   6n = 90

   n = 15

   Our first odd integer is 2n+1 = 2(15) + 1 = 31

   Our 3 consecutive odd integers are 31, 33, 35

 

Check:  31 +33+45 = 99  Yes

 

3)Which equation defines three consecutive odd integers that add up to 81?

A) n(n+1)(n+2)=81

B) n(n+2)(n+4)=81

C) n+(n+1)+(n+3)=81

D) n+(n+1)+(n+2)=81

E) n+(n+2)+(n+4)=81

 

We can immediately discard answers A, B because they are multiplying

instead of adding.

That leaves C, D, E

 

This question does not follow the logic I utilized in question 1 and 2. 

Assuming the first number is indeed odd and is called n

The next odd number is n+2 followed by n+4

Thus we must have:  n + (n+2) + (n+4) = 81

The answer is E

 

By the way:  The numbers are 25, 27, 29

5 0
3 years ago
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What are the solutions of the quadratic equation?<br> 5x^2+51x+54 = 0
pentagon [3]
X = - 6/5 or x=-9 !!!!!!!!!!!!!
7 0
3 years ago
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