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Lisa [10]
2 years ago
5

Charlotte measured the temperature at 7:00 A.M. and again at 3:00 P.M., She placed both measurements on a number line.

Mathematics
2 answers:
lions [1.4K]2 years ago
8 0
C I increased by 11 and 3/4
Margaret [11]2 years ago
6 0

Answer:

I think its C. It increased by 11 and 3/4 °C.

Step-by-step explanation:

I say that because it increases by 10 (-5 from +5) then to 6 which makes it 11 times. Then, it increases by 3/4 since its 2/4 + 1/4 which gives u 3/4. Also, its an increase and not a decrease because the numbers on the number line is going pass 0, so that shows an increase.

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Step-by-step explanation:

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Answer:

C) y = 2|x - 3| - 1

Step-by-step explanation:

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4 years ago
25% of the student's in Mrs.Brown's class like chocolate ice cream. How many students are in her class if 6 students like chocol
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Answer:

24 students.

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
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How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
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