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3 years ago

Please please please help me with this answer

Mathematics

1 answer:

kow [346]3 years ago

3 0

There are 3 girls .

9 students , 3 boys & 3 girls + another 3 boys because there are 3 more than girls so there are only 3 girls since you can't exceed 9 .

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Divide 6.391 Ã· 0.57. Round the quotient to the nearest hundredth.

Paul [167]

The answer is 11.21, dividing 6.391 by 0.57 will give you 11.2122807018.
Rounding to 11.21 because the thousandths place is less then 5 so the hundredths place stays the same.
(numeral)~(decimal)~(tenths)(hundredths)(thousandths)

5 0

3 years ago

bazaltina [42]

Answer:gotta go sorry

Step-by-step explanation:

3 0

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:

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

3 years ago

In ABC,if sin A=4/5 and tan A=4/3, then what I s cos A?

Svet_ta [14]

Cos(A) will be 3/5 Cos(A) is adjacent/hypotenuse. So here we have sin(A)=opp/hyp which is 4/5 and tan(A)=opp/adj which is 4/3. So cos(A) is adj./hyp and knowing the previous 2 terms of sin(A) and cos(A), you can go ahead and find Cos(A). So cos(A) is 3/5

5 0

3 years ago

DO NOT ANSWER IF YOU DO NOT KNOW

aliina [53]

Answer:

By saying "paying the same rate"

I believe you want to know for example when two people are buying a product how to know if they are paying the same amount.

If so then there are a couple of factors at play.

First of all you need to identify the GDP of the product and the amount of tax/VAT which if calculated properly may help you tell if they are paying the same amount or not.

3 0

3 years ago