Home / Expert Answers / Calculus / evaluate-int-left-6-x-3-5-x-2-5-x-4-right-d-x-int-left-6-x-3-5-x-2-5-x-4-right-pa818

(Solved): Evaluate. \[ \int\left(6 x^{3}+5 x^{2}-5 x+4\right) d x \] \[ \int\left(6 x^{3}+5 x^{2}-5 x+4\right ...





Evaluate.
\[
\int\left(6 x^{3}+5 x^{2}-5 x+4\right) d x
\]
\[
\int\left(6 x^{3}+5 x^{2}-5 x+4\right) d x=
\]
Evaluate.
\[
\int\left(\frac{5}{x}-7 e^{2 x}+e^{0.2}\right) d x
\]
\[
\int\left(\frac{5}{x}-7 e^{2 x}+e^{0.2}\right) d x=
\]
If the rate of excretion of a bio-chemical compound is given by
\[
f^{\prime}(t)=0.07 e^{-0.07 t} \text {, }
\]
the total amo
Under certain conditions, the number of diseased cells \( N(t) \) at time \( t \) increases at a rate \( N^{\prime}(t)=A e^{k
Under certain conditions, the number of diseased cells \( N(t) \) at time \( t \) increases at a rate \( N^{\prime}(t)=A e^{k
Evaluate. \[ \int\left(6 x^{3}+5 x^{2}-5 x+4\right) d x \] \[ \int\left(6 x^{3}+5 x^{2}-5 x+4\right) d x= \] Evaluate. \[ \int\left(\frac{5}{x}-7 e^{2 x}+e^{0.2}\right) d x \] \[ \int\left(\frac{5}{x}-7 e^{2 x}+e^{0.2}\right) d x= \] If the rate of excretion of a bio-chemical compound is given by \[ f^{\prime}(t)=0.07 e^{-0.07 t} \text {, } \] the total amount excreted by time \( \mathrm{t} \) (in minutes) is \( \mathrm{f}(\mathrm{t}) \). a. Find an expression for \( f(t) \). b. If 0 units are excreted at time \( t=0 \), how many units are excreted in 30 minutes? a. Find an expression for \( f(t) \). \[ f(t)= \] Under certain conditions, the number of diseased cells \( N(t) \) at time \( t \) increases at a rate \( N^{\prime}(t)=A e^{k t} \), where \( A \) is the rate of increase at time 0 (in cells per day) and \( \mathrm{k} \) is a constant. a. Suppose \( A=40 \), and at 5 days, the cells are growing at a rate of 120 per day. Find a formula for the number of cells after \( t \) days, given that 200 cells are present at \( t=0 \). b. Use your answer from part a to find the number of cells present after 11 days. a. Find a formula for the number of cells, \( N(t) \), after \( t \) days. \[ \mathrm{N}(\mathrm{t})= \] (Round any numbers in exponents to five decimal places. Round all other numbers to the nearest tenth.) Under certain conditions, the number of diseased cells \( N(t) \) at time \( t \) increases at a rate \( N^{\prime}(t)=A e^{k t} \), where \( A \) is the rate of increase at time \( O \) (in cells per day) and \( \mathrm{k} \) is a constant. a. Suppose \( A=40 \), and at 5 days, the cells are growing at a rate of 120 per day. Find a formula for the number of cells after \( t \) days, given that 200 cells are present at \( t=0 \). b. Use your answer from part a to find the number of cells present after 11 days. a. Find a formula for the number of cells, \( \mathrm{N}(\mathrm{t}) \), after \( \mathrm{t} \) days. \[ N(t)= \] (Round any numbers in exponents to five decimal places. Round all other numbers to the nearest tenth.)


We have an Answer from Expert

View Expert Answer

Expert Answer


We have an Answer from Expert

Buy This Answer $5

Place Order

We Provide Services Across The Globe