All categories

3 years ago

Rewrite in simplest terms: 10k-5(-6k+4)

Mathematics

1 answer:

Fittoniya [83]3 years ago

7 0

Answer:

Step-by-step explanation:

You might be interested in

-3y+1/3=-5/6y-5/2 what does y equal

kotykmax [81]

Answer:

17/13

Step-by-step explanation:

$- 3y + \frac{1}{3} = - \frac{5}{6}y - \frac{5}{2} \\ - \frac{18}{6}y + \frac{2}{6} = - \frac{5}{6} y - \frac{15}{6} \\ \frac{2}{6} + \frac{15}{6} = - \frac{5}{6} y + \frac{18}{6} y \\ \frac{17}{6} = \frac{13}{6} y \\ \frac{6}{13} \times \frac{17}{6} = y \\ \frac{17}{13} = y$

8 0

2 years ago

Which of the following correctly describes the interval shown?

adoni [48]

B is the correct one

5 0

3 years ago

if y is the midpoint of xz, y is located at (3,-1), and z is located at (11,-5), find the coordinates of x

svlad2 [7]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ X(\stackrel{x_1}{x}~,~\stackrel{y_1}{y})\qquad Z(\stackrel{x_2}{11}~,~\stackrel{y_2}{-5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{11+x}{2}~~,~~\cfrac{-5+y}{2} \right)=\stackrel{\stackrel{midpoint}{y}}{(3,-1)}\implies \begin{cases} \cfrac{11+x}{2}=3\\[1em] 11+x=6\\ \boxed{x=-5}\\ \cline{1-1} \cfrac{-5+y}{2}=-1\\[1em] -5+y=-2\\ \boxed{y=3} \end{cases}$

7 0

3 years ago

Read 2 more answers

Factor the polynomial F(x) =x^3-x^2-4x+4 completely.

Norma-Jean [14]

Answer:

Step-by-step explanation:

PART 1:

Possible roots by noticing the coefficient of first term:

x = 1, -1

PART 2:

1 | 1 -1 -4 4

1 0 -4

1 0 -4 0

The remainder is zero, hence one factor is (x-1)

PART 3:

$x^3-x^2-4x+4=(x^2-4)(x-1)\\\\x^3-x^2-4x+4=(x+2)(x-2)(x-1)$

PART 4:

$f(x)=(x+2)(x-2)(x-1)\\\\=(x^2-4)(x-1)\\\\=x^3-4x-x^2+4\\\\=x^3-x^2-4x+4$

6 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