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AnnyKZ [126]
3 years ago
8

If AB = 11, BC = 16 and CA = 14, list the angles of ABC in order from smallest to largest.

Mathematics
1 answer:
makvit [3.9K]3 years ago
5 0

Answer:

<h3>A,C,B 11,14,167#727272</h3>
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Which method would you use to prove that the two triangles are congruent?
Alona [7]

Answer:- AAS postulate


Explanation:-

  • AAS postulate tells that if two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.

Given:- One angle and one side of a triangle is equal to the one angle and one side of the other triangle.

We see there is one more pair of equal angles as they are vertically opposite angles . [See the attachment]

⇒ there is a triangle where two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.

⇒ The triangles are congruent [ by ASA postulate]


6 0
3 years ago
What is the solution to the system of equations?<br><br> y = 2x – 3.5<br><br> x – 2y = –14
iogann1982 [59]

Answer:

7

Step-by-step explanation:

Substitution then simplify

8 0
3 years ago
ALGEBRAIC EXPRESSION 11. Subtract the sum of 13x – 4y + 7z and – 6z + 6x + 3y from the sum of 6x – 4y – 4z and 2x + 4y – 7. 12.
Naily [24]

Answer:

Explained below.

Step-by-step explanation:

(11)

Subtract the sum of (13x - 4y + 7z) and (- 6z + 6x + 3y) from the sum of (6x - 4y - 4z) and (2x + 4y - 7z).

[(6x - 4y - 4z) +(2x + 4y - 7z)]-[(13x - 4y + 7z) + (- 6z + 6x + 3y) ]\\=[6x-4y-4z+2x+4y-7z]-[13x-4y+7z-6z+6x+3y]\\=6x-4y-4z+2x+4y-7z-13x+4y-7z+6z-6x-3y\\=(6x+2x-13x-6x)+(4y-4y+4y-3y)-(4z+7z+7z-6z)\\=-11x+y-12z

Thus, the final expression is (-11x + y - 12z).

(12)

From the sum of (x² + 3y² - 6xy), (2x² - y² + 8xy), (y² + 8) and (x² - 3xy) subtract (-3x² + 4y² - xy + x - y + 3).

[(x^{2} + 3y^{2} - 6xy)+(2x^{2} - y^{2} + 8xy)+(y^{2} + 8)+(x^{2} - 3xy)] - [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[x^{2} + 3y^{2} - 6xy+2x^{2} - y^{2} + 8xy+y^{2} + 8+x^{2} - 3xy]- [-3x^{2} + 4y^{2} - xy + x - y + 3]\\=[4x^{2}+3y^{2}-xy+8]-[-3x^{2} + 4y^{2} - xy + x - y + 3]\\=4x^{2}+3y^{2}-xy+8+3x^{2}-4y^{2}+xy-x+y-3\\=7x^{2}-y^{2}-x+y+5

Thus, the final expression is (7x² - y² - x + y + 5).

(13)

What should be subtracted from (x² – xy + y² – x + y + 3) to obtain (-x²+ 3y²- 4xy + 1)?

A=(x^{2} - xy + y^{2} - x + y + 3) - (-x^{2}+ 3y^{2}- 4xy + 1)\\=x^{2} - xy + y^{2} - x + y + 3 +x^{2}- 3y^{2}+ 4xy -1\\=2x^{2}-2y^{2}+3xy-x+y+2

Thus, the expression is (2x² - 2y² + 3xy - x + y + 2).

(14)

What should be added to (xy – 3yz + 4zx) to get (4xy – 3zx + 4yz + 7)?

A=(4xy-3zx + 4yz + 7)-(xy - 3yz + 4zx) \\=4xy-3zx + 4yz + 7 -xy + 3yz - 4zx\\=3xy-7zx+7yz+7

Thus, the expression is (3xy - 7zx + 7yz + 7).

(15)

How much is (x² − 2xy + 3y²) less than (2x² − 3y² + xy)?

A=(2x^{2} - 3y^{2} + xy)-(x^{2} - 2xy + 3y^{2})\\=2x^{2} - 3y^{2} + xy-x^{2} + 2xy - 3y^{2}\\=x^{2}-6y^{2}+3xy

Thus, the expression is (x² - 6y² + 3xy).

7 0
3 years ago
A: HL<br> B: ASA<br> C: SSS<br> D: SAS
evablogger [386]

Answer:

D: SAS

Step-by-step explanation:

6 0
3 years ago
|x| + 4 &gt; 7 plz help and hurry
Talja [164]

Answer:

The solution is (-\infty,-3)\cup(3,\infty).

Step-by-step explanation:

Given:

The inequality given is:

|x|+4>7

In order to simplify for 'x', we first isolate 'x' on one side.

Adding -4 on both sides, we get:

|x|+4-4>7-4\\\\|x|>3

Now, |x| is an absolute value function which is defined as:

|x|=\left \{ {{-x}\ \ x

Therefore, the given inequality can be rewritten as:

-x > 3\\x and x > 3

Therefore, the solution is (-\infty,-3)\cup(3,\infty).

4 0
3 years ago
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