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kakasveta [241]
3 years ago
5

Determine if they triangles are similar. If so, by what Theorem?

Mathematics
1 answer:
marusya05 [52]3 years ago
4 0

Answer:

SSS

Step-by-step explanation:

The sides of the smaller triangle equal the sides of the larger triangle when you multiply them by six, so they are parallel.

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ASAP PLS! What is 2k^2+3k-13=k^2+15?
shepuryov [24]

2k^2+3k-13=k^2+15

subtract k^2 -15 from each side

k^2 +3k -28=0

(k+7)(k-4)=0

k=-7  k=4

Answer: k= -7,4

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3 years ago
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mel-nik [20]

Answer:

x^2+10x+24

Step-by-step explanation:

(x+6)×(x+4)

=x^2+4x+6x+24

=x^2+10x+24 (ans)

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Your dividing number for both is by 8. Since 8 hours divided by 8 equals 1. You will do the same to the top and get an answer of 25 miles!!
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What times what equals -18
photoshop1234 [79]

-1 x 18 = 18

-2 x 9 = 18

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3 years ago
Solve the system:<br> -5x+6y=55<br> 4x+3y=34<br><br> Write down your solution.
svetlana [45]

Answer:

(1, 10)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Equality Properties

<u>Algebra I</u>

  • Solving systems of equations using substitution/elimination

Step-by-step explanation:

<u>Step 1: Define Systems</u>

-5x + 6y = 55

4x + 3y = 34

<u>Step 2: Rewrite Systems</u>

4x + 3y = 34

  1. Multiply everything by -2:                    -8x - 6y = -68

<u>Step 3: Redefine Systems</u>

-5x + 6y = 55

-8x - 6y = -68

<u>Step 4: Solve for </u><em><u>x</u></em>

<em>Elimination</em>

  1. Combine equations:                    -13x = -13
  2. Divide -13 on both sides:             x = 1

<u>Step 5: Solve for </u><em><u>y</u></em>

  1. Define equation:                    4x + 3y = 34
  2. Substitute in <em>x</em>:                       4(1) + 3y = 34
  3. Multiply:                                  4 + 3y = 34
  4. Isolate <em>y</em> term:                        3y = 30
  5. Isolate <em>y</em>:                                 y = 10
4 0
3 years ago
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