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navik [9.2K]
3 years ago
14

Can some one help me

Mathematics
2 answers:
Murljashka [212]3 years ago
7 0
The amswer would be 8.5 due to the length of the obeject
OlgaM077 [116]3 years ago
5 0

Answer:

see below

Step-by-step explanation:

10. CD = AC / 2 = 6 and CE = BC / 2 = 4.

11. DG = DH * 2.5 = 10

   HG = DG - DH = 10 - 4 = 6

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1. Look for a pattern in the first three equations. Then, fill in the missing numbers in the rest of the
Alla [95]

Answer:

1=1 times 2/2 1 + 2=2 times 3/2 1 +2+3=3 times 4/2 1+2+3+4=4 times 5/2 1+2+3+4+5=5 times 6/2

4 0
2 years ago
530 divided by 16 give the quotient and remainder
damaskus [11]
   
 
 
          33
      _____
16 | 530
       48
      -------
         50
         48
    ----------
            2

quotient = 33
remainder = 2








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2 years ago
Angle b and Angle g are known as alternate ___________ angles.
Afina-wow [57]
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4 0
3 years ago
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Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
How many days in january
sasho [114]
There are 31 days in january

7 0
2 years ago
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