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nataly862011 [7]
3 years ago
7

What triangles are similar to triangle AED? Explain or show your reasoning.

Mathematics
1 answer:
VLD [36.1K]3 years ago
4 0

Answer:

DCE is similar to AED

EBA is similar to AED

AFE is similar to AED

EFD is similar to AED

Step-by-step explanation:

Triangle AED:

D=53(alternate interior angle)

A=180-90-53=37 (angle of triangle)

AA similarity criterion

DCE is similar to AED

angle BEA=180-90-53=37=angle A of AED

Thus: AA similarity criterion:

EBA is similar to AED

The diagonal of parallelogram(here:rectangle) cut it into 2 congruent triangles

Thus:

EBA is congruent to AFE

EBA is congruent to AFEDCE is congruent to EFD

Congruent triangles are similar

Conclusion:

DCE is similar to AED

EBA is similar to AED

AFE is similar to AED

EFD is similar to AED

Please give brainliest if it helps

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128 pages+179 pages=307 pages
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One car is going 35mph the other car is going =55 mph one car started 18 min before the other car how long before the second car
Whitepunk [10]
First add the two numbers together 35 and 55 you will than get 90 take that and divide it by the number 18
ANSWER
90÷18=5
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3 years ago
Write a system of two equations in two variables where one equation is quadratic and the other is linear such that the system ha
PilotLPTM [1.2K]

Answer:

This system has no solution:

y = x²

y = 0 · x - 1

Step-by-step explanation:

Hi there!

I propose these two equations:

y = x²

y = 0 · x - 1

This system has no solution because in the quadratic function "y" is always positive for every value of "x", while in the linear function "y" is always -1 for every value of "x". Then, there will not be a pair (x,y) shared by both functions.

In other words, there is no value of "x" that satisfies this expression:

x² = 0 · x - 1

In the attached graphic, you can see that both curves never touch (in red is the quadratic function and in blue is the linear function), hence, the system has no solution.

Have a nice day!

4 0
4 years ago
An ellipse has vertices at (0, #17) and foci at (0, ±15). Write the equation of the ellipse in standard form. Graph the ellipse.
Lunna [17]

<u>ANSWER</u>

\frac{ {x}^{2} }{ 64 }  +  \frac{ {y}^{2} }{ 289 }  = 1

See attachment for the graph

<u>EXPLANATION</u>

The standard equation of the vertical ellipse with center at the origin is given by

\frac{ {x}^{2} }{ {b}^{2} }  +  \frac{ {y}^{2} }{ {a}^{2} }  = 1

where

{a}^{2}  \:  >  \:  {b}^{2}

The ellipse has its vertices at (0,±17).

This implies that:a=±17 or a²=289

The foci are located at (0,±15).

This implies that:c=±15 or c²=225

We use the following relation to find the value of b²

{a}^{2}  -  {b}^{2}  =  {c}^{2}

\implies \: 289  -  {b}^{2}  =  225

-  {b}^{2}  =  225 - 289

-  {b}^{2}  =  - 64

{b}^{2}  = 64

We substitute into the formula for the standard equation to get:

\frac{ {x}^{2} }{ 64 }  +  \frac{ {y}^{2} }{ 289 }  = 1

4 0
3 years ago
Write the slope-intercept form of the equation of the line given the
svetoff [14.1K]

Answer:

y= -4x-7 or 4x+y= -7

Step-by-step explanation:

we know y=mx+c

given,

m=-4 c=-7

the equation y= -4x-7 or 4x+y= -7

thank u

5 0
4 years ago
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