Consider the quadratic function f(y) = 8y2 – 7y + 6. What is the constant of the function?

1 answer:

8 0

The answer would be 6 because it has no variable. The formula for a quadratic function is f(x)=ax^2 +bx+ c, c which in this case is the constant because it has no variable.

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Find value of P and Q

ch4aika [34]

Answer:

$P= 6 \frac{1}{7}\\\\Q = 4 \frac{2}{7}$

Step-by-step Explanation

$\frac{5 + 3 \sqrt{2} }{5 - 3 \sqrt{2} } = P+Q\sqrt 2 \\ \\ \frac{(5 + 3 \sqrt{2}) }{(5 - 3 \sqrt{2}) } \times \frac{(5 + 3 \sqrt{2}) }{(5 + 3 \sqrt{2} )} = P+Q\sqrt 2 \\ \\ \frac{ {(5 + 3 \sqrt{2} )}^{2} }{ {(5)}^{2} - (3 \sqrt{2})^{2} } = P+Q\sqrt 2 \\ \\ \frac{ {5}^{2} + {(3 \sqrt{2} )^{2} + 2.5.3 \sqrt{2} } }{25 - 9 \times 2} = P+Q\sqrt 2 \\ \\\frac{ 25 + 18 + 30 \sqrt{2} }{25 - 18} = P+Q\sqrt 2 \\ \\\frac{ 43 + 30 \sqrt{2} }{7} = P+Q\sqrt 2 \\ \\ \frac{43}{7} + \frac{30 }{7} \sqrt{2} = P+Q\sqrt 2 \\ \\ equating \: like \: terms \: on \: both \: sides \\ \\ P = \frac{43}{7} \\ \\ \huge \red { \boxed{ P= 6 \frac{1}{7}}} \\ \\ Q = \frac{30}{7} \\ \\ \huge \orange { \boxed{ Q = 4 \frac{2}{7} }}$