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$\triangle DEF$ is the medial triangle of $\triangle ABC$ and $\triangle XYZ$ is the medial triangle of $\triangle DEF.$ If the

lbvjy [14]

A medial triangle is made from the midlines of the larger triangle, and its

length is half the length of the third side of the triangle.

The perimeter of ΔABC is 20

Reasons:

The medial triangle of triangle ΔABC is triangle ΔDEF

The medial triangle of ΔDEF is ΔXYZ

The perimeter of ΔXYZ = 5

By the midpoint theorem, we have;

The lengths of the sides of ΔDEF = 0.5 × The lengths of the sides of ΔABC

∴ The perimeter of ΔDEF = 0.5 × The perimeter of ΔABC

Which gives;

The perimeter of ΔABC = 2 × The perimeter of ΔDEF

Similarly, between ΔXYZ and ΔDEF, we have;

The perimeter of ΔDEF = 2 × The perimeter of ΔXYZ

∴ The perimeter of ΔDEF = 2 × 5 = 10

Therefore;

The perimeter of ΔABC = 2 × 10 = 20

The perimeter of ΔABC = 20

Learn more about the perimeter of a triangle here:

brainly.com/question/1340753

6 0

2 years ago

Which is true about the volume or surface area of these prisms?

GrogVix [38]

The correct answer is B- The surface area of B is greater than the surface area of A. I got this answer because...

Figure A's Volume= 60

Figure B's Volume= 60

Figure A's Surface Area= 104

Figure B's Surface Area= 248

So because the volumes are the same I knew it was not C or D, and then I solved the Surface Area for figures a and b and knew the correct answer is B.

Hope that helps. :)

3 0

3 years ago

Answeeeerr please ????

valentinak56 [21]

It is the first option.

4 0

2 years ago

Read 2 more answers

SOMEONE PLEASE HELP ME. WILL MARK BRAINLIEST

fredd [130]

Answer:

Function

Step-by-step explanation:

The following is a one-on-one function since neither the domain nor the range is being repeated.

6 0

2 years ago

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Suppose the graph of a cubic polynomial function has the same zeroes and passes through the coordinate (0, -5). Write the equati

amid [387]

In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
$f(x) = a(x - x_1)(x-x_2)(x-x_3)$

As we know zeros are that point where we will get value of function equal to zero. So it is basically in form $(x_n , 0)$

SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say $x_1 = 2 , x_2 = 3 , x_3 = 5$

So required equation is
$f(x) = a (x-2)(x-3)(x-5)$
              $= a[(x^2 - 2x - 3x + 6)(x-5)]$
              $= a[(x^2 - 5x+6)(x-5)]$
              $= a(x^3 - 5x^2 + 6x- 5x^2 + 25x - 30)$
              $= a(x^3 - 10x^2+31x-30)$
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
$-5= a (0-0+0 - 30)$
$-5 = -30a$
$a = \frac{-5}{-30} = \frac{1}{6}$
So required equation of cubic polynomial is
$f(x) = \frac{1}{6}(x^3 -10x^2+31x-30)$

For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.

4 0

3 years ago

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