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agasfer [191]
4 years ago
9

What are two ways to verify that your solutions are correct?

Mathematics
2 answers:
olga55 [171]4 years ago
5 0
The correct answers among the choices presented above are options B and D. Two ways to verify that solutions are correct are done by graphing the equations and see where they meet and by s<span>ubstitute back into both equations. The intersection of the graph signifies the solution. Also, when after substituting the solutions to both equations and they satisfy the equality then it is correct.</span>
Phantasy [73]4 years ago
3 0

Answer:

B. Graph the equations and see where they meet.

D. Substitute back into both equations.

Step-by-step explanation:

The two ways to verify that your solutions are correct are:

B. Graph the equations and see where they meet - Graphing is a very accurate and precise way to fetch solutions.

D. Substitute back into both equations - we can do this to cross check our solutions. This way makes sure that the solution is correct.

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BlackZzzverrR [31]

Answer:

see below

Step-by-step explanation:

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y = mx + b

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7 0
3 years ago
Please someone help me to prove this. ​
morpeh [17]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Use the Double Angle Identity: sin 2Ф = 2sinФ · cosФ

Use the Sum/Difference Identities:

sin(α + β) = sinα · cosβ + cosα · sinβ

cos(α - β) = cosα · cosβ + sinα · sinβ

Use the Unit circle to evaluate: sin45 = cos45 = √2/2

Use the Double Angle Identities:   sin2Ф = 2sinФ · cosФ

Use the Pythagorean Identity: cos²Ф + sin²Ф = 1

<u />

<u>Proof LHS → RHS</u>

LHS:                                  2sin(45 + 2A) · cos(45 - 2A)

Sum/Difference: 2 (sin45·cos2A + cos45·sin2A) (cos45·cos2A + sin45·sin2A)

Unit Circle:    2[(√2/2)cos2A + (√2/2)sin2A][(√2/2)cos2A +(√2/2)·sin2A)]  

Expand:        2[(1/2)cos²2A  + cos2A·sin2A + (1/2)sin²2A]

Distribute:              cos²2A   + 2cos2A·sin2A + sin²2A  

Pythagorean Identity:    1 + 2cos2A·sin2A

Double Angle:                1 + sin4A

LHS = RHS:  1 + sin4A = 1 + sin4A   \checkmark

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