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3 years ago

10

Gavin is building a model of the kitchen in the model of the floor tiles are withe 1/2 of the floor tiles are yellow and 1/10 of

the floor tiles are brown how many floor tiles could be in the model

1 answer:

jolli1 [7]3 years ago

7 0

2/5= 4 tiles
1/10= 1 tile
1/2= 5 tiles
Add them up and there is your answer: 10

*HOPE I HELPED!*

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If light travels at 10,000 km in 3.0 x 10² seconds,

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Answer:

1000xm

Step-by-step explanation:

1 meter = 3.2808 feet, hence. 9.8424 x 10^8 feet in 1 second. 1 foot in x seconds. hence it takes 1 / (9.8424 x 10^8) = 0.10168 x 10^(-8) seconds. ➡️1km = 1000xm⬅️

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Add.<br><br> Express your answer in simplest terms.<br><br> 1/5+2/3

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A company sells cookies in 250-gram packs. When a particular batch of 1,000 packs was weighed, the mean weight per pack was 255

AURORKA [14]

Let X be a random variable representing the weight of a pack of cookies.
P(X < 250) = P(z < (250 - 255)/2.5) = P(z < -5/2.5) = P(z < -2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275 = 2.3%

Therefore, we conclude that about 2.3% of the packs weighed less than 250 grams.

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3 years ago

What is a cubic polynomial function in standard form with zeros 1,1,and -3?

Mekhanik [1.2K]

Answer:

f(x) = x³ + x² - 5x + 3

Step-by-step explanation:

The cubic polynomial has zeros at 1, 1, - 3.

Therefore, x = 1, 1, - 3 are the roots of the polynomial and hence, (x - 1), (x - 1) and (x + 3) will be factors of the cubic polynomial.

Hence, we can write the polynomial as a function of x as

f(x) = (x - 1)(x - 1)(x + 3)

⇒ f(x) = (x² - 2x + 1)(x + 3)

⇒ f(x) = x³ - 2x² + 3x² + x - 6x + 3

⇒ f(x) = x³ + x² - 5x + 3

So, this is the cubic polynomial function in standard form. (Answer)

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3 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:

f(x) = cxⁿ
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

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3 years ago