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Anna007 [38]
2 years ago
8

. If CD = 4 and the perimeter of △ABC is 23, what is the perimeter of △ABE?

Mathematics
1 answer:
OlgaM077 [116]2 years ago
6 0

Answer:

31

Step-by-step explanation:

ABC has a perimeter of 23, so

   AB + BC + AC = 23

The perimeter of ABE would be the sum of these sides:

  AB + BC + CD + DE + AE

We also know:

  AE = AC, CD=4 and DE=4 because ACD is congruent to AED

So, AB + BC + CD + DE + AE = AB + BC + 4 + 4+ AC = 23+8 = 31

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Step-by-step explanation:

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Step-by-step explanation:

Hey there!!

Given,

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3 years ago
What are the roots of the equation? x^2(5x−7)(3x+2)=0
lisov135 [29]

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Whats the minimum and maximum value of f(x)=3(x+8)^2−10
gizmo_the_mogwai [7]

Answer:

(-8,-10)

Step-by-step explanation:

Rewrite (x+8)2(x+8)² as (x+8)(x+8).

f(x)=3((x+8)(x+8))−10

Expand (x+8) (x+8) using the FOIL Method.

Apply the distributive property.

f(x)=3(x(x+8)+8(x+8))−10

Apply the distributive property.

f(x)=3(x⋅x+x⋅8+8(x+8))−10
Apply the distributive property.

Simplify and combine like terms.

Simplify each term.

Multiply x by x.

f(x)=3(x2+x⋅8+8x+8⋅8)−10

Move 8 to the left of x.

f(x)=3(x2+8⋅x+8x+8⋅8)−10

Multiply 8 by 8.

f(x)=3(x2+8x+8x+64)−10

Add 8x and 8x.

f(x)=3(x2+16x+64)−10

Apply the distributive property.

f(x)=3x2+3(16x)+3⋅64−10

Simplify.

Multiply 16 by 3.

f(x)=3x2+48x+3⋅64−10

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f(x)=3x2+48x+192−10

Subtract 10 from 192.

f(x)=3x2+48x+182

The minimum of a quadratic function occurs at x=-\frac{b}{2a} If a is positive, the minimum value of the function is f (-\frac{b}{2a}).

Substitute in the values of aa and b.

x=−\frac{48}{2(3)}

x=-8

Replace the variable x with −8 in the expression.

f(−8)=3(−8)2+48(−8)+182

Y=-10

Therefore, the minimum value is (-8,-10) but if it is asking for just the y-value it would be -10.

7 0
2 years ago
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