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hammer [34]
3 years ago
8

Describe some possible combinations (at least 2) of 4-seat tables and 6-seat tables that will seat 160 customers. Explain how yo

u found them. Write an equation to represent the situation. What do the variables represent?
Mathematics
1 answer:
MrMuchimi3 years ago
6 0

Answer:

You could do 50 2 seat tables and 5 4 seat tables

You could do you could do 40 2 seat tables and 10 4 seat tables

Step-by-step explanation:

brainliest plz

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Simplify the following expression.
Blababa [14]

Answer:

-6x

Step-by-step explanation:

I just answered this

it will equal -6x

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5 0
3 years ago
Jensen has an above ground pool in the shape of a regular octagon. Opposite sides are 25 feet apart and each side is 10.4 feet l
bija089 [108]

Answer:

<u>Given</u>: base = 10.4ft, height = 12.5

Area of octagon = 8(1/2 × b × h)

  • 8(1/2 × 10.4 × 12.5)
  • 520ft²

Volume of pool = 520ft² × 3ft

  • 1560ft³

Now, 1 cubic ft takes 7.5 gallons to fill.

Therefore, 1560 cubic ft takes,

  • 1560 × 7.5
  • 11700

So, <u>Correct choice</u> - [D] 11,700.

6 0
3 years ago
May someone please help with the way please
Oksana_A [137]

Answer:

17.6 m²

Step-by-step explanation:

Given the ratio of similar shapes = a : b, then

area of shapes = a ² : b²

Δ PTQ and Δ PRS are similar and so the ratio of corresponding sides are equal, that is

PT : PR = 6 : 9 = 2 : 3, thus

ratio of areas = 2² : 3² = 4 : 9

let the area of Δ PQT be x, then using proportion

\frac{4}{x} = \frac{9}{x+22} ( cross- multiply )

9x = 4(x + 22) ← distribute

9x = 4x + 88 ( subtract 4x from both sides )

5x = 88 ( divide both sides by 5 )

x = 17.6

Thus area of Δ PQT = 17.6 m²

5 0
3 years ago
.The Tlingit hunted seals in which culture region?
MrRa [10]
C. Pacific Northwest
4 0
2 years ago
<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7
salantis [7]

Answer:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

Substitute in <em>x</em> = 0 once more:

\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0
3 years ago
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