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3 years ago

Where is error hellp

Mathematics

2 answers:

Lemur [1.5K]3 years ago

8 0

Answer: Step 2

Step-by-step explanation:

Note: Every positive number has two square roots: a positive square root and a negative square root. Step 2 should have been x+4 = ±11.

xeze [42]3 years ago

3 0

Answer:

Step 2

Step-by-step explanation:

Every positive number has two square roots.

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My Notes A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 f

Sergio [31]

Answer:

6.667 ft/s

Step-by-step explanation:

Given

Height of street light = 15ft

Height of man = 6ft

Speed away from pole = 4ft/s

Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.

This forms a similar triangle (see attachment below).

From similar triangles, we have

(y - x)/6 = y/15 ----- Solve equation

15(y - x) = 6 * y

15y - 15x = 6y ---- Collect Like Terms

15y - 6y = 15x

9y = 15x --- divide through by 9

y = 15x/9

y = 5x/3 ---- differentiate with respect to time, t

dy/dt = 5/3 dx/dt

dx/dt represents rate of change of distance per time = 4ft/s

while dy/dt represents rate of movement of his shadow tips

dy/dt = 5/3 * 4

dy/dt = 20/3 = 6.667 ft/s

4 0

2 years ago

Read 2 more answers

Pls help I need help plsss

cupoosta [38]

Whats up? What do you need help with

4 0

3 years ago

g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for <em>h: </em>

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for <em>h</em>.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$