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3 years ago

Can you answer this plz

Mathematics

1 answer:

Sedaia [141]3 years ago

3 0

Go on google and then type in calculator for long division and you'll find the answers and the work

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What is the value of d?

blondinia [14]

Answer:

its 82 or 83 but u should use a protractor

ep-by-step explanation:

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3 years ago

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25% of 68000 is how much

hjlf

$1 \% =\frac{1}{100} \\ \\25 \% =\frac{25}{100}=0,25\\ \\25 \% \cdot 68000 = 0,25 \cdot 68000 =17 000$

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2 years ago

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PLEASE HELP I NEED THIS ASAP Find the vertex of y=x^2-7x+4

Goshia [24]

Answer:

$\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

Step-by-step explanation:

$y=x^2-7x+4\\\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}\\\mathrm{The\:parabola\:params\:are:}\\a=1,\:b=-7,\:c=4\\x_v=-\frac{b}{2a}\\x_v=-\frac{\left(-7\right)}{2\cdot \:1}\\\mathrm{Simplify}\\x_v=\frac{7}{2}\\\mathrm{Plug\:in}\:\:x_v=\frac{7}{2}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}\\y_v=\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4\\$

$\mathrm{Simplify\:}\left(\frac{7}{2}\right)^2-7\cdot \frac{7}{2}+4:\quad -\frac{33}{4}\\y_v=-\frac{33}{4}\\Therefore\:the\:parabola\:vertex\:is\\\left(\frac{7}{2},\:-\frac{33}{4}\right)\\\mathrm{If}\:a0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}\\a=1\\\mathrm{Minimum}\space\left(\frac{7}{2},\:-\frac{33}{4}\right)$

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3 years ago

People that are smart in GEOMETRY then answer this PLZ

Viefleur [7K]

Answer:

11. Corresponding

12. Alternate Interior

13. Consecutive Interior

14. Corresponding

15. Alternate Exterior

16. Alternate Interior

8 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

2 years ago