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Ainat [17]
3 years ago
5

How do you solve the factoring of 3t^3-7t^2-3t+7

Mathematics
1 answer:
Serjik [45]3 years ago
7 0

Step-by-step explanation:

<h3>t^2(3t-7) -1(3t-7)</h3>

(3t-7) (t^2-1)this is the answer

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4x-8=12 when x=-2<br> verify the solution of the equation and explain.
mojhsa [17]

Answer:

Solve in terms of the arbitrary variable x . x = 5 5 = − 2

Step-by-step explanation:

7 0
3 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
A scuba diver descends 0.75 meters per second. What is her change in elevation after 25 seconds?
vesna_86 [32]

Answer:

-18.75 Meters

Step-by-step explanation:

0.75 x -25 = -18.75 Meters :D

4 0
3 years ago
Find the unit rate<br><br> 8 hits in 22 games
Aleksandr [31]

Answer:

0.363636

Step-by-step explanation:

Hits per game is hits / games

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8 0
3 years ago
Side XY of triangle XYZ is extended to point W, creating a linear pair with &lt;WYZ and &lt;XYZ. What is the value of x?​
bulgar [2K]

Answer:

80

Step-by-step explanation:

I just did this and its right!

3 0
3 years ago
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