Please help with this question.

During a scuba dive, Lainey descended to a point 19 feet below the ocean surface. She continued her descent at a rate of 25 feet

Leno4ka [110]

Answer:

$25t+19\leq 144$

$t\leq5$

The number of minutes she can continue to descend if she does not want to reach a point more than 144 feet below the ocean surface is <u>at most 5 minutes.</u>

Step-by-step explanation:

Given:

Initial depth of the scuba dive = 19 ft

Rate of descent = 25 ft/min

Maximum depth to be reached = 144 ft

Now, after 't' minutes, the depth reached by the scuba dive is equal to the sum of the initial depth and the depth covered in 't' minutes moving at the given rate.

Framing in equation form, we get:

Total depth = Initial Depth + Rate of descent × Time

Total depth = $19+25t$

Now, as per question, the total depth should not be more than 144 feet. So,

$\textrm{Total depth}\leq 144\ ft\\\\19+25t\leq 144\\\\or\ 25t+19\leq 144$

Solving the above inequality for time 't', we get:

$25t+19\leq 144\\\\25t\leq 144-19\\\\25t\leq 125\\\\t\leq \frac{125}{25}\\\\t\leq 5\ min$

Therefore, the number of minutes she can continue to descend if she does not want to reach a point more than 144 feet below the ocean surface is at most 5 minutes.

7 0

3 years ago

Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means: