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iragen [17]
3 years ago
6

jingle bell jingle bell jingle bell rockJingle bell, jingle bell, jingle bell rock Jingle bells swing and jingle bells ring Snow

in' and blowin' up bushels of fun Now the jingle hop has begun Jingle bell, jingle bell, jingle bell rock Jingle bells chime in jingle bell time Dancin' and prancin' in Jingle Bell Square In the frosty air What a bright time, it's the right time To rock the night away Jingle bell time is a swell time To go glidin' in a one-horse sleigh Giddy-up jingle horse, pick up your feet Jingle around the clock Mix and a-mingle in the jinglin' feet That's the jingle bell rock Jingle bell, jingle bell, jingle bell rock Jingle bells chime in jingle bell time Dancin' and prancin' in Jingle Bell Square In the frosty air What a bright time, it's the right time To rock the night away Jingle bell time is a swell time To go glidin' in a one-horse sleigh Giddy-​
Mathematics
2 answers:
Kobotan [32]3 years ago
7 0
Not me singing this in my head now currently dancing in my room as if it is Christmas.
earnstyle [38]3 years ago
3 0

Answer:

GAS STATION SUSHI

Step-by-step explanation:

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The polynomial f(x) is written in factored form:
Ray Of Light [21]

Set each set of parentheses to 0 and solve for x.

X—6 = 0

X = 6

X + 5 = 0

X = -5

X-9 = 0

X = 9

The zeros are 6, -5, 9

7 0
3 years ago
Write an equation in slope-intercept form of the line that passes through (−5, 19) and (5, 13)
kotykmax [81]
13-19/5+5=-6/10=-3/5
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Evaluate a+4 when a =7
LenaWriter [7]

If a = 7 then a+4 = 11.

a+4 is basically 7+4.

Hope this helps :)

6 0
2 years ago
Read 2 more answers
Please help as soon as possibly with this math question will give brainlyest
vovikov84 [41]
This is the answer for ur question

5 0
2 years ago
A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
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