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Mathematics

1 answer:

sesenic [268]4 years ago

8 0

$f(x) = {(x + \frac{b}{2})}^{2} + \frac{24 - b}{2}$

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Y=(5x-2)(2x+3) intercept and vertex

ahrayia [7]

2x=3 this is what I got

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3 years ago

Match the expression on the left with the correct simplified expression on the right.

andriy [413]

Given: The expression below

$\begin{gathered} (\frac{(3x^3y^4)^3}{(3x^2y^2)^2})^2 \\ (\frac{(3x^4y^2)^4}{(3x^5y^2)^3})^2 \end{gathered}$

To Determine: The matching expression to the given expressions

Solution

Let us simplify each of the expressions using exponents rule

$\begin{gathered} Exponent-Rule1=(a^m)^n=a^{m\times n} \\ Exponent-Rule2=(\frac{a^m}{a^n})=a^{m-n} \end{gathered}$

Applying the exponent rule 1 above to the given expressions

$\begin{gathered} (3x^3y^4)^3=3^3x^{3\times3}y^{4\times3}=27x^9y^{12} \\ (3x^2y^2)^2=3^2x^{2\times2}y^{2\times2}=9x^4y^4 \end{gathered}$ $\begin{gathered} (3x^4y^2)^4=3^4x^{4\times4}y^{2\times4}=81x^{16}y^8 \\ (3x^5y^2)^3=3^3x^{5\times3}y^{2\times3}=27x^{15}y^6 \end{gathered}$

Applying the exponent rule 2

$\frac{(3x^{3}y^{4})^{3}}{(3x^{2}y^{2})^{2}}=\frac{27x^9y^{12}}{9x^4y^4}=\frac{27}{9}x^{9-4}y^{12-4}=3x^5y^8$ $\frac{(3x^{4}y^{2})^{4}}{(3x^{5}y^{2})^{3}}=\frac{81x^{16}y^8}{27x^{15}y^6}=\frac{81}{27}x^{16-15}y^{8-6}=3xy^2$

Let us not apply exponent rule 1 above

$(\frac{(3x^{3}y^{4})^{3}}{(3x^{2}y^{2})^{2}})^2=(3x^5y^8)^2=3^2x^{5\times2}y^{8\times2}=9x^{10}y^{16}$ $(\frac{(3x^{4}y^{2})^{4}}{(3x^{5}y^{2})^{3}})^2=(3xy^2)^2=3^2x^2y^{2\times2}=9x^2y^4$

Hence, the matching is as shown below