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12345 [234]
3 years ago
8

Can anyone do question 25 part c? :)

Mathematics
1 answer:
inna [77]3 years ago
5 0
\begin{cases}x^3\qquad\qquad\ \ \  when\ x\geq3\\5\qquad\qquad\quad \ when\  0\ \textless \ x\ \textless \ 3\\x^2-x+2\quad\ when\ x\leq0 \end{cases}\\\\\\ (a)\\ f(0)\iff x=0\implies f(x)=x^2-x+2\\\\f(0)=0^2-0+2=2\\\\(b)\\ \{f(2)\iff x=2;\ \ f(1)\iff x=1\}\implies f(x)=5\\\\ f(2)=5\quad\wedge\quad f(1)=5\\\\ f(2)-f(1)=5-5=0

(c)\\ \big\forall \limits_{ n\in R}\ n^2\geq 0\ \ \Rightarrow\ \ \big\forall \limits_{ n\in R}\ (-n^2)\leq 0 \ \ \Rightarrow\ \ x\leq0\ \ \Rightarrow\ \  f(x)=x^2-x+2\\\\f(-n^2)=(-n^2)^2-(-n^2)+2=n^4+n^2+2
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Answer:

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56 = −8d + 8 there's not multiple answers
ollegr [7]

Answer:

d= -6

Step-by-step explanation:

56 = -8d + 8

1) You want to get d by itself. Subtract 8 from both sides.

56 = -8d + 8

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<em>(8 - 8 = 0, and 56 - 8 is 48.)</em>

48 = -8d

2) Then, you divide my -8 on both sides.

(<em>48 / -8 is -6, and -8d / -8 is just d.)</em>

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4 years ago
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marshall27 [118]

Answer:

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Step-by-step explanation:

Step 1:

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Step 2:

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Step 3:

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Answer:

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