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elixir [45]
4 years ago
7

What is the minimum width of the square base?

Mathematics
1 answer:
riadik2000 [5.3K]4 years ago
4 0
4 I think I might be wrong so sorry if I am wrong
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Please solve I need help!
ladessa [460]
24 is the answer welcome !
5 0
3 years ago
Solve x/3 = x+2/2. X=
Leni [432]

Answer:

I will assume that the term "x+2/2" is meant to be "(x + 2)/2)."  Otherwise the equation would read (x/3) = x + 1

Step-by-step explanation:

(x/3) = (x + 2)/2

x = 3*(x+2)/2   [Multiply both sides by 3]

x = (3x + 6)/2

x = (3/2)x + 6/2

x - (3/2)x = 3

-0.5x = 3

x = -6

====================

Check:

Does (x/3) = (x + 2)/2 for x = -6?

(-6/3) = (-6 + 2)/2

-2 = -4/2

-2 = -2  YES

6 0
2 years ago
Aster sold 18 oranges if this are 12% of her total oranges how many oranges are not sold​
WITCHER [35]

Answer:

Step-by-step explanation:

So 18 oranges are 12 percent of her total oranges. You can use Ratio to solve this.

O : P

18 : 12                               P = Percentage     O = Orange

1.5 : 1

150 : 100                          

All I did was make the percentage one by dividing both numbers by 12. This makes it 1.5 oranges per percent. Times 1.5 with 100 to make it 100 percent.

7 0
3 years ago
Read 2 more answers
Make a list of factors for -20
andre [41]

Answer:

1 2 4 5 10 20

-1 -2 -4 -5 -10 -20

7 0
4 years ago
Read 2 more answers
Cos (90-theta) • cosec90-theta) =tano. How?​
denis-greek [22]

Answer:

<u>______________________________________________________</u>

<u>TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-</u>

For any right angled triangle with one angle α ,

  • \cos (90 - \alpha  ) = \sin \alpha  or  \sin(90 - \alpha ) = \cos\alpha
  • cosec \: (90 - \alpha  ) = \sec\alpha   or  \sec(90 - \alpha ) = cosec\:\alpha

<u>SOME GENERAL TRIGNOMETRIC FORMULAS :-</u>

  • <u></u>\sin \alpha = \frac{1}{cosec \: \alpha }  or  cosec \: \alpha  = \frac{1}{\sin \alpha }
  • <u></u>\cos \alpha = \frac{1}{\sec \alpha }  or  \sec \alpha = \frac{1}{\cos \alpha }

<u>______________________________________________________</u>

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

\cos (90 - \alpha ) \times cosec(90 - \alpha )

=> sin\alpha  \times \sec\alpha

=> \sin\alpha  \times \frac{1}{\cos\alpha }

=> \frac{\sin\alpha }{\cos\alpha }

=> \tan\alpha = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

3 0
3 years ago
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