Prove that ΔABC and ΔEDC are similar. ABC and DEC where angles A and E are right angles, AC equals 4, AB equals 3, BC equals 5, DC equals 15, DE equals 9, and CE equals 12 15 over 5 equals 12 over 4 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate. ∠DCE is congruent to ∠CBA by the Vertical Angles Theorem and 15 over 5 equals 12 over 4 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate. ∠E and ∠B are right angles and, therefore, congruent since all right angles are congruent. 9 over 4 and 12 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate. 15 over 4 equals 12 over 5 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate.

We can prove they are similar by proving that triangle EDC is 3 times the size of triangle ABC. (see explanation for proof).

Step-by-step explanation:

Looking at triangle ABC, we can see that the side lengths are 3, 4, and 5.

Let's look at the hypotenuse for now.

The hypotenuse of ABC is 5.

Looking at triangle EDC, the side lengths are 9, 12, and 15.

Again, looking at the hypotenuse, it's 15 for EDC.

To find how much bigger the hypotenuse of EDC is then ABC, let's divide.

Now, these triangles will only be similar if EVERY SINGLE SIDE of ABC × 3 is equal to the sides of EDC.

The width of ABC is 3.

The width of EDC is 9.

so this works.

The length of ABC is 4.

The length of EDC is 12.

So we know that these triangles are similar because the side lengths of EDC are 3 times the amount of the side lengths of ABC.

Hope this helped!

the answer is c.

step-by-step explanation:

given : the shape of his satellite can be modeled by   where x and y are modeled in inches.

now,the given equation   is a equation of parabola.

here, coefficient of x is positive, hence the parabola opens rightwards.

on comparing this equation with standard equation , we get

in standard equation, coordinates of focus=(0,m)

thus for given equation coordinates of focus=(0,1.5)

step-by-step explanation: