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3 years ago

9

PLEASE CAN SOMEONE HELP

2 answers:

xxTIMURxx [149]3 years ago

7 0

2.1. 5a + 2b
2.2. -4x
2.3. 3x + 5
2.4. 10y

Sergio [31]3 years ago

5 0

Answer:

2.1) 5a + 2b

2.2) -4x

2.3) 3x + 5

2.4) 14y

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

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

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3 years ago

Given the diagram, a supplement to

Likurg_2 [28]

Given:

The figure of a quadrilateral ABCD.

To find:

The supplement of angle ABC.

Solution:

We know that, two angles are called supplementary angles if their sum is 180 degrees.

$\angle ABC+\angle BAC=110^\circ+40^\circ$

$\angle ABC+\angle BAC=150^\circ\neq 180^\circ$

So, $\angle BAC$ is not a supplement of $\angle ABC$ .

$\angle ABC+\angle ADC=110^\circ+90^\circ$

$\angle ABC+\angle ADC=200^\circ$

So, $\angle ADC$ is not a supplement of $\angle ABC$ .

$\angle ABC+\angle DAB=110^\circ+(40+30)^\circ$

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$\angle ABC+\angle DAB=180^\circ$

So, $\angle DAB$ is a supplement of $\angle ABC$ .

In the same way $\angle ABC+\angle BDC\neq 180^\circ$ .

So, $\angle BDC$ is not a supplement of $\angle ABC$ .

Therefore, the correct option is C.

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Step-by-step explanation:

So lets count the time between 10:53 and 1:05

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2 Hours to one

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Total 2 hours 12 mins divided by 2 to find the middle

1 hour 6 mins. We add 1 hr 6 mins to 10:53.

10:53 + 6mins equals 11:59

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Answer c

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Answer:

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Step-by-step explanation:

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