3 years ago

Are the triangles similar? If so, why?

Mathematics

1 answer:

Julli [10]3 years ago

4 0

Answer:

they are not similar because they have different size

You might be interested in

A typical cell phone plan has a fixed base fee that includes a certain amount of data and an overage charge for data use beyond

AnnyKZ [126]

Answer:

The equation will be:

C(g) = 15g + 62, if g > 2

8 0

3 years ago

A certain human red blood cell has a diameter of 0.000007 meters. Which expression represents this diameter, in meters, in scien

vazorg [7]

Answer:

Step-by-step explanation:

3 0

4 years ago

X 8. From the given Venn-diagram, list the elements of the following sets, a) U b) XU Y and XUY c) X Y and X d) X - Y and X-Y e)

const2013 [10]

Answer:

where is the Venn diagram plz insert Venn diagram then I should solve

6 0

3 years ago

The fact that financial services are more plentiful and easier to access means that

Virty [35]

<span> Not 100% sure but I think it is: consumers need to be better informed about what services are available and what those services cost.</span>

8 0

3 years ago

PLEASE HELP!!

erik [133]

$2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha$

$1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}$

$3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x$

8 0

3 years ago