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1 year ago

28.4.2 Test (CST): Transformations, Congruence, and Similarities, Part 2

Mathematics

1 answer:

frez [133]1 year ago

8 0

Answer:

R'(2,4)

Step-by-step explanation:

When doing translations you subtract or add within the coordinates like a normal problem.

If it helps you, write it as seperate equations like:

1-1 and 5-1

Hope that helps ✓

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Gala2k [10]

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Please help me on this will give you brainliest!!

dolphi86 [110]

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Step-by-step explanation:

since the triangles are similar, the sides are in a ratio

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aev [14]

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2 years ago

If the given is DEFG is a kite how do I prove that triangle EFH is congruent to triangle GFH

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1 year ago

Verify that this trigonometric equation is an identity?

Vlada [557]

$\cot x\sec^4 x=\cot x+2\tan x+\tan^3x\\\\L=\dfrac{\cos x}{\sin x}\cdot\dfrac{1}{\cos^4x}=\dfrac{1}{\sin x}\cdot\dfrac{1}{\cos^3x}=\dfrac{1}{\sin x\cos^3x}\\\\R=\dfrac{\cos x}{\sin x}+2\cdot\dfrac{\sin x}{\cos x}+\dfrac{\sin^3x}{\cos^3x}\\\\=\dfrac{\cos x\cos^3x}{\sin x\cos^3x}+\dfrac{2\sin x\cos^2x}{\cos x\sin x\cos^2x}+\dfrac{\sin^3x\sin x}{\cos^3x\sin x}\\\\=\dfrac{\cos^4x+2\sin^2x\cos^2x+\sin^4x}{\sin x\cos^3x}\\\\=\dfrac{(\cos^2x)^2+2\sin^2x\cos^2x+(\sin^2x)^2}{\sin x\cos^3x}$

$=\dfrac{(\cos^2x+\sin^2x)^2}{\sin x\cos^3x}=\dfrac{1}{\sin x\cos^3x}=L\\\\Used:\\\tan(a)=\dfrac{\sin(a)}{\cos(a)}\\\cot(a)=\dfrac{\cos(a)}{\sin(a)}\\\sec(a)=\dfrac{1}{\cos(a)}\\\sin^2a+\cos^2a=1\\(a+b)^2=a^2+2ab+b^2$

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3 years ago