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3 years ago

To prove that DEF congruent DGF by SAS,what additional information is needed

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

8 0

Answer:

∠DFE ≅ ∠DFG

Step-by-step explanation:

we know that

Triangles are congruent by SAS, if any pair of corresponding sides and their included angles are equal in both triangles

In this problem we have that

The pair of corresponding sides that are equal are

EF=GF and DF (is the same side for both triangles)

The included angle are ∠DFE and ∠DFG

therefore

The additional information needed to prove that triangles DEF and DGF are congruent by SAS is that angles ∠DFE and ∠DFG are equal

∠DFE ≅ ∠DFG

kompoz [17]3 years ago

7 0

Answer:de=dg

Step-by-step explanation:

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Maddie and her friends are making s'mores over the campfire. they have a bag of 48 marshmallows, 2 packs of graham crackers cont

VladimirAG [237]

The answer would be 16 S'mores and the limiting reactant would be the grahams.

(This is assuming that S'mores would need 2 grahams, 1 marshmallow and 1 chocolate piece.)

Limiting reactant would be the reactant that runs out first.

Let's take your problem into account and see what we have:

48 marshallows

32 grahams (2 x 16 per pack)

45 chocolate pieces (5 x 15 pieces per bar)

Since need 2 of the grahams per S'more then the maximum yield of the grahams is 16 S'mores.

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3 years ago

Consider the parent function f(x)=e^x and transformed function g(x)=-f(x)-4. Which features of function F function G are differe

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3 years ago

There are three clubs at Redwood High School: Debate, Student council, and Key club. There are 1000 students in the school. And

hram777 [196]

Answer:

The number of students in the key club is 490

Step-by-step explanation:

The given parameters are;

The number of student in the school = 1000

The total number of student in the debate club = 310

The total number of student in the student council = 650

The total number of student who are in debate and student council = 170

The total number of student who are in both debate and the key club = 150

The total number of student who are in both student council and the key club = 180

The number of students who are in all three clubs = 50

Therefore, we have;

Let A represent the number of students in the debate club

Let B represent the number of students in the student council

Let C represent the number of students in the key club

We have;

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

Where;

n(A∪B∪C) = 1000

n(A) = 310

n(B) = 650

n(A∩B) = 170

n(B∩C) = 180

n(C∩A) = 150

n(A∩B∩C) = 50

Therefore;

n(C) = n(A∪B∪C) - (n(A) + n(B) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C))

Substituting the values gives;

n(C) = 1000 - (310 + 650 - 170 - 180 - 150 + 50) = 490

The number of students in the key club, n(C) = 490.

5 0

3 years ago