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2 years ago

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The function c(x) = 3x − 4 determines how many cupcakes need to be purchased for a birthday party, where x is the number of kids

at the party. The party planner uses d(c(x)) to find the amount of money to bring for the cupcake purchase. The function d(x) = 2x + 5. Solve for how much money to bring when there are 7 kids at the party.

2 answers:

dezoksy [38]2 years ago

6 0

The answer of d(c(x))=39. hope it will help you

Karolina [17]2 years ago

5 0

Answer: $39

Step-by-step explanation:

Given: The function $c(x)=3x-4$ determines how many cupcakes need to be purchased for a birthday party, where x is the number of kids at the party. The party planner uses d(c(x)) to find the amount of money to bring for the cupcake purchase. The function $d(x) = 2x + 5$

Now, $d(c(x))=2(3x-4)+5=6x-8+5=6x-3$

To find the amount of money to bring when there are 7 kids at the party, put x in d(c(x)), we get

$d(c(7))=6(7)-3=42-3=39$

Hence, the amount of money to bring when there are 7 kids at the party = $39

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Solve for <em>x</em> by dividing boths ides by 5 log(4) - log(3) :

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3 years ago

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Since, f(xy) ≠ f(x)f(y)

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Part B:

Given $f:Z_2 \rightarrow Z_2,$ defined by $f(x)=-x$

Note that in $Z_2$ , -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular $f(x)=x$

$f(x+y)=x+y \\ \\ f(x)+f(y)=x+y$

and

$f(xy)=xy \\ \\ f(x)\cdot f(y)=xy$

Therefore, the function is a homomorphism.

Part C:

Given $g:Q\rightarrow Q$ , defined by $g(x)= \frac{1}{x^2+1}$

$g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}$

Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.

Part D:

Given $h:R\rightarrow M(R)$ , defined by $h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)$

$h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)$

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$h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)$

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Given $f:Z_{12}\rightarrow Z_4$ , defined by $\left([x_{12}]\right)=[x_4]$ , where $[u_n]$ denotes the lass of the integer $u$ in $Z_n$ .

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and

$f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)$

Therefore, the function is a homomorphism.

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3 years ago

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Alecsey [184]

Hello,

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2 years ago

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Nookie1986 [14]

Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

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Left to Right

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<u>Algebra I</u>

Terms/Coefficients

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

(-5x + 6x²) - (6x² - 5x)

<u>Step 2: Simplify</u>

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3 years ago

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