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3 years ago

9

Use the formulas to find the lateral area and the surface area of the given prism round your answer to the nearest whole number

5.39m 26m 5m 2m
the choices for the answer is
322m2 327m2
296m2 342m2
296m2 332m2
322m2 332m2

2 answers:

just olya [345]3 years ago

8 0

The Answer is 296m2 342m2

soldi70 [24.7K]3 years ago

3 0

The answer is 322 m^2

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Answer:

The shopper should but the 6 pcs pack because its cheaper.

Step-by-step explanation:

6-pcs pack = $2.10

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There are two friends. Each friend has a dollar to buy some pie. The first friend buys two-eighths of a pie for a dollar. The se

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Answer:

Neither they get the same amount

Step-by-step explanation:

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3 years ago

In a unit circle, ø = 270º. Identify the terminal point and sin ø.

larisa [96]

Answer:

Terminal point (0, -1); sin Ø = -1 ⇒ A

Step-by-step explanation:

In the unit circle, Ф is the angle between the terminal side and the positive part of the x-xis

The terminal point on the positive part of the x-axis is (1, 0),which means Ф = 0° or 360° and cosФ = 1, sinФ = 0

The terminal point on the positive part of the y-axis is (0, 1),which means Ф = 90° and cosФ = 0, sinФ = 1

The terminal point on the negative part of the x-axis is (-1, 0),which means Ф = 180° and cosФ = -1, sinФ = 0

The terminal point on the negative part of the y-axis is (0, -1),which means Ф = 270° and cosФ = 0, sinФ = -1

In a unit circle

∵ Ф = 270°

→ By using the 4th rule above

∴ The terminal point is (0, -1)

∴ sinФ = -1

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4 0

2 years ago

Read 2 more answers

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

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2 years ago

Read 2 more answers

A man is walking down a very long flight of

Alex

Answer:

33.6 seconds

Step-by-step explanation:

if he begins at 136 feet and wants to go to 42 feet, the distance is 94 feet

d = 94

r = 2.8

d = rt; solving this formula for 't':

t = d/r

t = 94/2.8

t = 33.6 seconds

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3 years ago