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vladimir2022 [97]

3 years ago

15

Pls decipher the code. Will give brainliest for correct answer tyyyyy

1 answer:

Karo-lina-s [1.5K]3 years ago

5 0

Answer:

M has a library card number divisible by three

Step-by-step explanation:

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If you get it right, you get a brainliness!

S_A_V [24]

Answer:

b

Step-by-step explanation:

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3 years ago

If you take 3 by 2 what do you get

jekas [21]

By multiplying the factors together you’ll get 6

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4 years ago

Need help I’ll give brainliest if right

givi [52]

It’s c
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4 0

3 years ago

Calculate the slope of the following points. (0, 4) and (5, 9). Show all work.

Volgvan

Answer:

1

Step-by-step explanation:

Subtract the y values and put them in the numerator (top of fraction). 9-4=5.

Subtract the x values and put them in the denominator (bottom) 5-0=5.

Put the numbers in the fraction and simplify. 5/5 = 1.

5 0

3 years ago

Read 2 more answers

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

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

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

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

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

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago