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nevsk [136]
3 years ago
14

PLEASEE SOMEONE ANSWER ILL DO ANYTHING ANYTHING PLEAASEEE HELP ME this is myh 6th time posting this ;[

Mathematics
1 answer:
Sonbull [250]3 years ago
4 0
What is the problem?
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We are given AB ∥ DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal
ohaa [14]

Answer: AAA similarity.


Step-by-step explanation:  CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have

∠CED ≅ ∠CBA,

∠CDE ≅ ∠CAB

and

∠DCE ≅ ∠ACB [same angle]

Hence, by AAA (angle-angle-angle) similarity,

△CED ~ △ABC.

Thus, the correct option is AAA similarity.


8 0
3 years ago
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Hi, so first u have to multiple 12×6=72 then next u divide by 2 which must five u a food start
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2 years ago
Send help<br><br><br> djkhkjdhkhjksdhkh
Sedaia [141]

Answer:

B

Step-by-step explanation:

I hope this helps you out!

3 0
2 years ago
Reagan is using geometry software to construct a regular pentagon by using rotations.
LekaFEV [45]
I would assume that he's using a 90° rotation.
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3 years ago
Find the dy/dx for the equation below using implicit differentiation.
n200080 [17]

First of all, recall that y is a product of function of x. We have three factors:

f(x)=x^2,\quad g(x)=e^{3x},\quad h(x)=\sin(4x)

The derivative of a product of function is computed by deriving one function at the time, and then adding all the results:

y' = f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)

Let's compute the derivative of each function first:

f'(x)=2x,\quad g'(x) = 3e^{3x},\quad h'(x)=4\cos(4x)

Now plug f, f', g, g', h, h' in the formula above as required:

2xe^{3x}\sin(4x)+3x^2e^{3x}\sin(4x)+4x^2e^{3x}\cos(4x)

6 0
3 years ago
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