All categories

3 years ago

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

You might be interested in

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

6 0

2 years ago

Angle b and Angle g are known as alternate ___________ angles.

Afina-wow [57]

Alternate exterior angles because the are OUTSIDE of the parallel lines

4 0

3 years ago

Read 2 more answers

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

Read 2 more answers