Hello! I can help you with this! Do -75/5 to get -15. -15 will be your middle number. So for consecutive integers, -13 + -14 + -15 + -16 + -17 = 75. Those are five consecutive integers that have a sum of 75. The lowest of 5 integers is -17, because it’s the smallest of them all and the furthest to the left of the number line. A line further to the left of the number line means it’s smaller. Therefore, the rest of the five integers is -17.

6 0

3 years ago

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Please help differentiate this!!!!!!!!!

vlada-n [284]

$\bf h=20ln(3t+2)+30\\\\
-------------------------------\\\\
\boxed{a}\\\\
\stackrel{0~years}{t=0}\qquad h=20ln[3(0)+2]+30\implies h=20ln(2)+30
\\\\\\
h\approx 43.86
\\\\\\
\boxed{b}\\\\
\stackrel{1~meter}{h=100}\qquad 100=20ln(3t+2)+30\implies 70=20ln(3t+2)
\\\\\\
\cfrac{70}{20}=ln(3t+2)\implies \stackrel{\textit{log cancellation rule}}{e^{\frac{7}{2}}=e^{ln(3t+2)}}\implies e^{\frac{7}{2}}=3t+2
\\\\\\
e^{\frac{7}{2}}-2=3t\implies \cfrac{e^{\frac{7}{2}}-2}{3}=t\implies 10.371817\approx t$

$\bf \boxed{c}\\\\
\cfrac{dh}{dt}=20\left(\cfrac{1}{3t+2}\cdot 3 \right)+0\implies \cfrac{dh}{dt}=20\left(\cfrac{3}{3t+2} \right)\\\\\\ \cfrac{dh}{dt}=\cfrac{60}{3t+2}
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\left. \cfrac{dh}{dt} \right|_{3}\implies \cfrac{60}{3(3)+2}\implies \cfrac{60}{11}
\\\\\\
\left. \cfrac{dh}{dt} \right|_{10}\implies \cfrac{60}{3(10)+2}\implies \cfrac{15}{8}$

5 0

3 years ago

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