Suppose ab=ae can you use the sss postulate or the sas postulate to prove abc= aed ?

Option 4 is correct that is both apply

Step-by-step explanation:

We have given  the triangle we have to tell which postulate SSS or SAS  to use to prove

ΔABC=ΔAED

We can use both of them

Case1: Since, three of the sides are equal that is

AB=AE

AC=AD

BC=ED

Which means SSS can be used

Since, SSS is side side side

Case2: Since one angle and two sides are equal

AB=AE

AC=AD

And ∠BAC=∠EAD

Which means SAS can be used

Since, SAS is side angle side

Therefore, Option 4 is correct that is both apply.

 both apply

Step-by-step explanation:

SAS congruence postulate says that if two sides and the included angle of a triangle are congruent to two sides and the included angle of other triangle then the two triangles are said to be congruent.

In the given triangles ΔABC and ΔAED , we have

∠BAC ≅ ∠EAD

AC ≅ AD

BE ≅ DE

If AB ≅ AE , then we have sufficient things to proof that ΔABC ≅ ΔAED by SSS congruence postulate .

i.e. for AC ≅ AD , BE ≅ DE and AB ≅ AE   [all three sides are congruent]

ΔABC ≅ ΔAED  by SSS congruence postulate.

Also, If AB ≅ AE , then we have sufficient things to proof that ΔABC ≅ ΔAED by SAS congruence postulate .

i.e. for AC ≅ AD   [Side]

∠BAC ≅ ∠EAD    [included angle]

AB ≅ AE             [Side]

⇒ ΔABC ≅ ΔAED  by SAS congruence postulate.

Hence, we can apply both postulates to prove triangles congruent .

4y - 6 + y^2 - 9 = y^2 + 4y - 15

y^2 term

terms

- terms

total of 3 terms when simplified

step-by-step explanation:

(a) x intercepts: -6 and 2

you can obtain these by solving the equation 0=x^2 +4x-12 using your favorite quadratic formula. but it is easy to verify -6 and 2 indeed satisfy this equation.

(b) y intercept: -12

this is obtained by setting x=0: 0^2+4*0-12 = -12

(c) the minimum value is: -16 (this is the vertex)

you can get the vertex by rewriting the equation into the form:

there is no global or local maximum as the function goes to infinity, so only in the limit sense: +infinity.