All categories

2 years ago

I NEED HELP AHHH-HHHHHHHHH HELP ME ASAP

Mathematics

1 answer:

zvonat [6]2 years ago

4 0

Answer:

1/5, 1/3, 4/6

Step-by-step explanation:

First, find all the common denominators for these numbers. In this case, it's 30.

1/5 = 10/30

4/6 = 20/30

1/5 = 6/30

Now, you can see which ones are greatest and least.

1. 1/5 (least)

2. 1/3 (middle)

3. 4/6 (greatest)

You might be interested in

Jeffrey plans to rent a beach house for sister's birthday and will use the savings to pay for it the company charges to $25 clea

balu736 [363]

He will be using $85 for all of that

5 0

2 years ago

Solve for x:2/5 (x - 2) = 4x

Helen [10]

Answer:

x= -2/9

Step-by-step explanation:

2(x-2)/5=4x

2(x-2)=4x*5

2(x-2)=20x

x-2=20x/2

x-2=10x

-2=10x-x

9x= -2

x= -2/9

7 0

2 years ago

450 g in the ratio 2:13

jonny [76]

450 g in 2:13

Total proportion asked for =2+13=15

Now comparing g with ratio, or whatever

450/15=30g $\implies$ Therefore 1 in ratio depicts of 30g

So final answer = $\boxed{60g : 390g}$

$\huge\mathfrak\colorbox{white}{}$

More to know, daily bit of knowledge: The Binomial Theorem:

$\boxed{{(a+b)} ^n= \sum_{k=0}^{n}\binom{n}{k} a^{n-k}b^{k}}$

3 0

2 years ago

Susan was given the math problem x - 4 + -11. She found the answer to be 7. Critique her answer. Is she correct? How do you know

Dmitrij [34]

Answer:

she was wrong

Step-by-step explanation:

x-4-11 = 0 (4+-1=4-1)

x=15

hence she was wrong

3 0

2 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