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Vedmedyk [2.9K]
3 years ago
11

Find the missing factor B that makes the equality true42x^5y^4=(-7x^3y)(B)​

Mathematics
1 answer:
Ksju [112]3 years ago
4 0

Answer:

-6x^2 y^3 = B

Step-by-step explanation:

42x^5y^4=(-7x^3y)(B)​

Solve for B

Divide by -7

42x^5y^4/-7=(-7x^3y)(B)/-7

-6 x^5 y^4 = x^3 y B

Divide by x^3

-6 x^5 y^4/x^3 = x^3 y B​/x^3

-6 x^5/x^3 y^4 = By

-6 x^2 y^4

Divide by y

-6x^2 y^4/y = By/y

-6x^2 y^3 = B

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In Exercises 9-14, decide whether enough information
Degger [83]

The triangles that can be proved to be congruent by the SAS Congruence Theorem are:

10. ΔLMN and ΔNQP

13. ΔEFH and ΔGHF

The triangles that cannot be proved to be congruent by the SAS Congruence Theorem due to insufficient information are:

9. ΔABD and ΔCDB

12. ΔQRV and ΔTSU

14. ΔKLM and ΔMNK

11. ΔYXZ and ΔWXZ

<h3>What is the SAS Congruence Theorem?</h3>
  • If two triangles have two pairs of corresponding sides that are congruent, and a pair of corresponding included angle that are congruent, both triangles can be proven to be congruent triangles by the SAS Congruence Theorem.

Thus:

9. ΔABD and ΔCDB have:

  • one pair of congruent <em>non-included angles </em>- ∠ABD ≅ ∠CDB
  • two pairs of congruent sides - AD ≅ BC, and BD ≅ BD

ΔABD and ΔCDB lack a pair of <em>included angles</em>, therefore, the information given is not enough to prove the triangles are congruent by the SAS Congruence Theorem.

10. ΔLMN and ΔNQP have:

  • one pair of congruent <em>included angles </em>- ∠LMN ≅ ∠NQP
  • two pairs of congruent sides - LM ≅ NQ, and MN ≅ QP

ΔLMN and ΔNQP, therefore, have enough information to prove that they are congruent triangles by the SAS Congruence Theorem.

11. ΔYXZ and ΔWXZ have:

  • one pair of congruent <em>non-included angles </em>- ∠YXZ ≅ ∠WXZ
  • two pairs of congruent sides - XZ ≅ XZ, and XW ≅ YZ

ΔYXZ and ΔWXZ lack a pair of <em>included angles</em>, therefore, the information given is not enough to prove the triangles are congruent by the SAS Congruence Theorem.

12. ΔQRV and ΔTSU do not have enough information to prove they are congruent by the SAS Congruence Theorem.

13. ΔEFH and ΔGHF have two pairs of congruent sides and a pair of congruent <em>included angles</em><em>, </em>therefore, there is enough information to prove they are congruent by the SAS Congruence Theorem.

14. ΔKLM and ΔMNK do not have enough information to prove they are congruent by the SAS Congruence Theorem.

Therefore, the triangles that can be proved to be congruent by the SAS Congruence Theorem are:

10. ΔLMN and ΔNQP

13. ΔEFH and ΔGHF

The triangles that cannot be proved to be congruent by the SAS Congruence Theorem due to insufficient information are:

9. ΔABD and ΔCDB

12. ΔQRV and ΔTSU

14. ΔKLM and ΔMNK

11. ΔYXZ and ΔWXZ

Learn more about SAS Congruence Theorem on:

brainly.com/question/13408604

3 0
3 years ago
How to convert decimal to binary by hand?
Katena32 [7]
You have to divide decimal number by 2 and then to write the reminder. Then you have to continue to divide the result by 2 and so on. At the end you have to read all those reminders ( numbers 0 and 1 ) backwards and you will get the binary number. Example: Convert 10 ( decimal ) to binary.
10 : 2 = 5  |   0
  5 : 2 = 2  |   1
  2 : 2 = 1  |   0
  1 : 2 = 0  |   1
                        ↑
Finally:  10 ( decimal ) = 1010 ( binary )
7 0
3 years ago
Simplify the expression 2x+4(3y+x).
inysia [295]

Answer:

The answer in simplified form would be 6x+12y.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What do open circles mean
Virty [35]

Answer: according to mathbitshomework.com: “When graphing a linear inequality on a number line, use an open circle for "less than" or "greater than", and a closed circle for "less than or equal to" or "greater than or equal to". ... The solution set for this problem will be the full graph of both inequalities, since the two inequalities do not overlap.”

6 0
3 years ago
Kevin caught some fish. Of them 4/9 were herring, and 2/9 salmon. What was the total amount of fish, amount of herring and amoun
julia-pushkina [17]

There were 3 different fish caught, Salmon, Herring and Flounder.

4/9 were Herring, 2/9 were Salmon, 4/9 +2/9 = 6/9

This means 3/9 were Flounders ( 3/9 + 6/9 = 9/9 = 1)

3/9 can be reduced to 1/3.

1/3 of the fish were Flounders.

Divide the amount of flounders by the portion caught:

12 / 1/3 = 12 * 3/1 = 36 total fish.

Now you have total number of fish, multiply the total by each portion for each type of fish.

Total fish = 36

Herring = 4/9 x 36 = 16

Salmon = 2/9 x 36 = 8

6 0
3 years ago
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