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3 years ago

Determine the measures of the angles of AMNO if the measures of the angles are in the ratio 2:4:6.

Mathematics

1 answer:

masha68 [24]3 years ago

8 0

Answer:

60,120,180

Step-by-step explanation:

360:(2+4+6)=30

30×2=60

30×4=120

30×6=180

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3 years ago

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3 years ago

Kylie explained that (–4x + 9)2 will result in a difference of squares because (–4x + 9)2 = (–4x)2 + (9)2 = 16x2 + 81. Which sta

OverLord2011 [107]

Well, in order to get that you had to do distributive property, distributive property is when you get the outside number next to the parenthesis and multiply it by the numbers or variable inside the parenthesis one by one. So if she was trying to explain how you got the answer you got then you would say you used the distributive property. Now the (-4x)2 would look like (-4x)(-4x) nad the (9)2 will look like (9)(9) that is why it results in 16x2+81. When you square something it means that you multiply the first number or variable or both times the same. That is why you got that answer. Now I am going to make clear that you r(-4x) changed from a negative to a postive because when you multiply a negative times a negative(or divide also) it results in a positve.

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8 0

3 years ago

Read 2 more answers

Find the sum of the series and show your process for finding each term leading to the final sum.

Crazy boy [7]

ANSWER

$\sum_{k=3}^5( - 2k + 5) = - 9$

EXPLANATION

The given series is

$\sum_{k=3}^5( - 2k + 5)$

This series is finite.

The expanded form is

$\sum_{k=3}^5( - 2k + 5) = ( - 2 \times 3 + 5) + ( - 2 \times 4 + 5) + ( - 2 \times 5 + 5)$

This simplifies to

$\sum_{k=3}^5( - 2k + 5) = ( - 6 + 5) + ( - 8+ 5) + ( - 10+ 5)$

$\sum_{k=3}^5( - 2k + 5) = ( - 1) + ( - 3) + ( - 5)$

$\sum_{k=3}^5( - 2k + 5) = - 9$

5 0

3 years ago

Prove : (sec θ - tan θ )^2 = 1 - sin θ /1+sin θ

labwork [276]

$(sec x - tan x)^2 \\ \\ = sec^2x - 2 sec x tan x + tan^2 x \\ \\ =(1+tan^2 x) - 2 sec x tan x +tan^2 x \\ \\ =1 - 2 sec x tan x + 2 tan^2 x \\ \\ = 1 - 2tan x(sec x - tan x) \\ \\ =1 - \frac{2 sin x}{cos x} (\frac{1-sin x}{cos x}) \\ \\ = 1 - \frac{2 sin x (1-sin x)}{cos^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{1-sin^2 x} \\ \\ =1 - \frac{2 sin x (1-sin x)}{(1-sin x)(1+sin x)} \\ \\ =1-\frac{2 sin x}{1+sin x}$

3 0

3 years ago