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3 years ago

Gcf 60x^4y^7, 45x^5y^5 and 75x^3y

1 answer:

6 0

Steps:

What do all of these have in common?
~ all of the leading coefficients are divisible by 5
~ all have at least 3 x's
~ all have at least 1 y

Now all we have to do is pull those out:

answer: 5x^3y

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If at the beginning of july Minh sweeps up 80 grams of sand each day how much will she have at the end of july

Annette [7]

Answer: 2,480 grams of sand

Step-by-step explanation:

There are 31 days in July.

Minh sweeps up 80 grams every day in July so at the end of July, Minh should have:

= Number of days * Sand per day

= 31 * 80

= 2,480 grams of sand

8 0

3 years ago

How do you subtract 1 integer by another

nikdorinn [45]

Use absolute value. If both signs are negative you will get a negative number. if both are positive you get positive. if the signs are different, subtract the smaller absolute value from the larger absolute value.

8 0

3 years ago

Read 2 more answers

<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:

Calculus

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 x = 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 x = 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 x = 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

2 years ago

Do you guys know 1.2n-1.5=0.45n

Lera25 [3.4K]

Answer:

n = 2

Step-by-step explanation:

1.2n - 1.5 = 0.45n

collect like terms

1.2n - 0.45n = 1.5

0.75n = 1.5

divide both side by 0.75

0.75n/0.75 = 1.5/0.75

n =2

4 0

2 years ago

What is the domain? What is the range?

almond37 [142]

Answer:

The domain of a function is the set of all possible inputs for the function. For example, the domain of f(x)=x² is all real numbers, and the domain of g(x)=1/x is all real numbers except for x=0. We can also define special functions whose domains are more limited.

The range is the difference between the largest and smallest numbers. The mid range is the average of the largest and smallest number.

Hope this helps!

8 0

2 years ago