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3 years ago

who can do my algebra hw there are just 4 problems :) will give BRAINLIST. btw i need your sna. or ig

Mathematics

2 answers:

Simora [160]3 years ago

5 0

Answer:

Ok if u answer my questio...

Step-by-step explanation:

Firdavs [7]3 years ago

5 0

i can my snp is lit_dogface

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Tamara purchased 31 cases of tile to use on her bathroom floors .she used 17 cases in her upstairs bathroom and the rest in her

Vikki [24]

The correct equation would be 17 + x = 31

6 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

3 years ago

HCF of smallest 2 digit composite number and smallest 3 digit prime number is?

posledela

Answer:

HCF = 1

Step-by-step explanation:

Smallest 2 digit composite number = 10

Smallest 3 digit prime number = 101

Factors of 10 = 1 , 2 , 5

Factors of 101 = 1 , 101

HCF = 1

6 0

2 years ago

X + 4y = 18 3x − y = 2 given x + 4y = 18 12x − 4y = 8 x + 4y + 8 = 18 + 8 x + 4y + (12x − 4y) = 18 + 8 13x = 26 justification st

goldenfox [79]

Spacing needs to be more clear instead of having a giant paragraph. you do not write math in giant paragraphs do you

5 0

3 years ago

9th grade ..Plz solve this maths

8_murik_8 [283]

since the polygon has 8 chambers the angle of one chamber = 360/8 = 45°

so the angle WOP = 45°

now drawing a imaginary line which seperates the triangleWOP into half

so the angle of imaginary line is 45/2 = 22.5

sin(theta) = opp/hyp

since WP (opp) is unknown. take as x

sin(22.5) = x/44

0.383×4 = x

x = 1.532 (app) = 1.5 cm

perimeter = 1.5 × 8

= 12 cm

area of triangle WOP = 2×1/2 ×bh = bh

= 4×1.5

= 6 cm²

so the area of regular polygon = 8× area of a triangle

= 48 cm²

6 0

3 years ago

who can do my algebra hw there are just 4 problems :) will give BRAINLIST. btw i need your sna. or ig​

who can do my algebra hw there are just 4 problems :) will give BRAINLIST. btw i need your sna. or ig