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elena-14-01-66 [18.8K]

1 year ago

10

The order for points only matter when naming a ..?

2 answers:

Gnoma [55]1 year ago

5 0

Answer:

Co-ordinate (x, y)

Step-by-step explanation:

ZanzabumX [31]1 year ago

5 0

Answer:

what are you saying??? explain more for answers

Step-by-step explanation:

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Graph the image of the given triangle under a dilation with a scale factor of and center of dilation (0, 0)

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2 years ago

Solve for x. <br> 35x = 21

emmasim [6.3K]

X = 21/35

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Explanation: divide both sides by 35 and reduce the fraction 21/35 to get 3/5.

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2 years ago

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F(x)=-3x-5 and g(x) =4x-2 find (f-g)(x)

Triss [41]

<span>F(x)=-3x-5 and g(x) =4x-2 find (f-g)(x)

</span>(f-g)(x) = -3x-5 - (4x-2)
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2 years ago

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Ivahew [28]

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2 years ago

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

2 years ago