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Vsevolod [243]
2 years ago
8

(80 pts) please help 1 easy question # 5 !! Please please

Mathematics
2 answers:
MaRussiya [10]2 years ago
7 0
The third one (f+g)(x)=4x6-2x2-5x+8
babunello [35]2 years ago
6 0

Step-by-step explanation:

#5

(f+g) (x) = x³-2x²+1 + 4x³-5x+7

= 5x³-2x²-5x+8 (option D)

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\bold{\rm{Given}}\text{ : If y - 3x = 9, 7x + y = 25, what is the value of x and y}?

\bold{\rm{Solution}} \text{: We'll solve this by substitution method}

\dashrightarrow  y - 3x = 9 \\  \\  \dashrightarrow   \boxed{y = 9 + 3x } \qquad .. \sf eq(1)

\text{we got the value of y now getting the value of x}

\looparrowright 7x + y = 25 \\  \\  \looparrowright 7x = 25 - y \\  \\  \looparrowright  \boxed{x =   \frac{25 - y}{7}} \qquad .. \sf eq(2)

\text{Now, putting y = 9 + 3x in eq(2)}

\hookrightarrow  x =  \frac{25 - y}{7}   \\  \\ \hookrightarrow  x =  \frac{25 - 9 + 3x}{7}  \\  \\ \hookrightarrow  7x =  25 - 9 + 3x \\  \\ \hookrightarrow  7x - 3x =  16 \\  \\ \hookrightarrow  4x = 16 \\  \\ \hookrightarrow  x =  \frac{16}{4}  \\  \\ \star \quad  \boxed{ \green{ \frak{ x = 4}}}

\text{Now we know the value of x, for getting y we need to put x in eq(1)}

: \implies y = 9 + 3x \\  \\   : \implies y = 9 + 3(4) \\  \\  : \implies y = 9 + 12  \\  \\    \star \quad  \boxed{\frak{  \green{y = 21}}}

\text{Hence, values of x and y are} \:  \frak{ \green{4}}  \: \text{and} \:  \frak{ \green{21}} \: \text{respectively}.

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