All categories

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

You might be interested in

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

3 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