3 years ago

The first decision is a choice between

Mathematics

2 answers:

Setler79 [48]3 years ago

5 0

D. min-mid-max-mid-min

C. -cosine

pantera1 [17]3 years ago

4 0

Answer:

min-mid-max-mid-min

-cosine

Step-by-step explanation:

This is the correct answer, further proof in the file attached.

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How do you do <br> (x+y)(x-5y)+3(x+y)^2<br> ???? <br> Emergency, please help with full steps

Naddik [55]

Answer:

$4 {x}^{2} + 2xy - 2 {y}^{2}$

Step-by-step explanation:

$(x + y)(x - 5y) + {3(x + y)}^{2} \\ x(x - 5y) + y(x - 5y) + 3(x + y)(x + y) \\ {x}^{2} - 5xy + xy - 5 {y}^{2} + (3x + 3y)(x + y) \\ {x}^{2} - 4xy - 5 {y}^{2} + 3x(x + y) + 3y(x + y) \\ {x}^{2} - 4xy - 5 {y}^{2} + 3 {x}^{2} + 3xy + 3xy + 3 {y}^{2} \\ {x}^{2} - 4xy - 5 {y}^{2} + 3 {x}^{2} + 6xy + 3 {y}^{2} \\ {x}^{2} + 3 {x}^{2} - 4xy + 6xy - 5 {y}^{2} + 3 {y}^{2} \\ 4 {x}^{2} + 2xy - 2 {y}^{2}$