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3 years ago

15

A label is placed around the juice box below, not including the bases. Determine the surface area of the label

1 answer:

nasty-shy [4]3 years ago

8 0

Answer:

3 In²

Step-by-step explanation:

Talking about surface area, the total area of an object is taken into consideration.

NB: Kindly check the attached picture for the diagram of the 3D object.

In this question, the surface area has to be determined and we will have to take into consideration all the area of the 3D object given in the question.

Surface Area = 6 × k².

Surface area = 6 × ( 1.8 × 1.8) cm².

Surface Area = 19.44 cm².

The next thing to do here is to convert the cm² to in². Therefore, 19.44 cm² ------> x in².

19.44 × 0.155 in² = 3 in².

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A patio in the shape of a regular octagon (8-sided figure) has a side length of 24 ft and an apothem length of 60 ft. What is th

Damm [24]

Given:
Octagon-shaped patio
Side length = 24ft
apothem length = 60ft

Area of a regular octagon = 2 * a² * (1 + √2)
a = side length

A = 2 * (24ft)² * (1 + √2)
A = 2 * 576ft² * (1 + 1.414)
A = 2 * 576ft² * 2.414
A = 2,780.93 ft²

8 0

3 years ago

Mr. Ray wanted to measure the height of a light post. He placed a mirror on the ground 32 feet from the light post, then walked

ryzh [129]

Answer:

Option (2) is the answer.

Step-by-step explanation:

From the figure attached,

AB represents the length of Mr. Ray, ED represents the light post and the mirror was placed at point C.

Here AB = d = 6 feet

CD = w = 32 feet

BC = f = 9 feet

Since both the triangles ABC and EDC are similar, their corresponding sides will be in the same ratio.

$\frac{AB}{DE}=\frac{BC}{CD}$

$\frac{6}{h}=\frac{9}{32}$

h = $\frac{6\times 32}{9}$

h = 21.33 feet

h ≈ 21.3 feet

Therefore, height of the light post is 21.3 feet.

Option (2) is the answer.

7 0

3 years ago

What is the slope of the line that passes through (1,-4) and (-4,-6)

Fudgin [204]

The slope of the line is 4/10

remember: rise over run

6 0

3 years ago

A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t

11Alexandr11 [23.1K]

Answer:

<u>Question 1:</u> $dm/dt=0.2m$ <u />

<u />

<u>Question 2:</u> $m=Ae^{(0.2t)}$ <u />

<u />

<u>Question 3:</u> $m=10e^{(0.2t)}$ <u />

<u />

<u>Question 4:</u> $m=10g$ <u />

Explanation:

<u>Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m</u>

<u></u>

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

<u>Question 2: Find the general solution of this equation. Use A as a constant of integration.</u>

a) <u>Separate variables</u>

$dm/m=0.2dt$

b)<u> Integrate</u>

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) <u>Antilogarithm</u>

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

<u>Question 3. Which particular solution matches the additional information?</u>

<u></u>

Use the measured rate of 4 grams per hour after 3 hours

$t=3hours,dm/dt=4g/h$

First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

<u>Particular solution:</u>

$m=10e^{(0.2t)}$

<u>Question 4. What was the mass of the bacteria at time =0?</u>

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

3 0

3 years ago

Plz help ASAP!!!!!!!!

lilavasa [31]

Answer:

The answer is 95.03

Step-by-step explanation

The formula that I used

A=πr^2

Area = π·5.5^2 ≈ 95.03318

Hope that helps :)

4 0

3 years ago