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2 years ago

Solve the equation-2/3x = -12 show work plz

Mathematics

2 answers:

Marina CMI [18]2 years ago

4 0

Answer:

x=1/18

Step-by-step explanation:

equation-2/3x = -12

-2=-12×3x

-2=-36x

x=-2/-36=1/18

Lorico [155]2 years ago

3 0

Answer:

Step-by-step explanation:

-2/3x = -12

-2 = 3x × -12

-2 = -36x

2 = 36x

2/36 = x

1/18 = x

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SOMEONE HELP ME PLEASE

jenyasd209 [6]

Answer: The median, because the data distribution is skewed to the left

EXPLANATION

Given the box plot with the following parameters:

Minimum value at 11

First Quartile, Q1 at 22.5

Median at 34.5

Third Quartile, Q3 at 36

Maximum value at 37.5

First, we notice that the data distribution is skewed to the left because the median (34.5) is closer to the third quartile (36) than to the first quartile

(22.5).

Furthermore, we know that the mean provides a better description of the center when the data distribution is symmetrical while the median provides a better description of the center when the data distribution is skewed.

Therefore, we conclude that for the given box plot, the median will provide a better description of the center because the data distribution is skewed to the left.

4 0

3 years ago

Find the exact values of all three trigonometric functions for each angle.

My name is Ann [436]

Answer:

Sin (270º)= -1

Cos (270º) = 0

Tan (270º) = -∞

Sin (330º) = -0.5

Cos (330º) = $\frac{\sqrt{3} }{2}$ = 0.8660

Tan (330º) = - $\frac{\sqrt{3} }{3} \\$ = -0.57735

Step-by-step explanation:

3 0

3 years ago

GEW has angle measures given by m/_G=5x + 66, m?/_ E = 3x, and m/_ W= 142 -4x whats the mesure of e

8090 [49]

Answer:

21°

Step-by-step explanation:

GEW has angle measures given by m/_G=5x + 66, m?/_ E = 3x, and m/_ W= 142 -4x whats the mesure of e

= 5x + 66 + 3x + 142 - 4x = 180

4x + 208 = 180

4x = 208 - 180

4x = 28

x = 28/4

x = 7

Solving for E = 3x

E = 3 × 7

E = 21°

7 0

2 years ago

Evaluate the expression:

FinnZ [79.3K]

Answer:

vw= 22

Step-by-step explanation:

To find the dot product of vw, multiply the corresponding numbers and add them.

v= <3, -8, -3> w= <-4, -2, -6>

vw= (3*-4)+(-8*-2)+(-3*-6)

vw= -12+16+18

vw= 4+18

vw= 22

6 0

3 years ago

Hello again! This is another Calculus question to be explained.

podryga [215]

Answer:

See explanation.

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions

Function Notation

Exponential Property [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

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 following and are trying to find the second derivative at x = 2:

$\displaystyle f(2) = 2$

$\displaystyle \frac{dy}{dx} = 6\sqrt{x^2 + 3y^2}$

We can differentiate the 1st derivative to obtain the 2nd derivative. Let's start by rewriting the 1st derivative:

$\displaystyle \frac{dy}{dx} = 6(x^2 + 3y^2)^\big{\frac{1}{2}}$

When we differentiate this, we must follow the Chain Rule: $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]$

Use the Basic Power Rule:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} (2x + 6yy')$

We know that y' is the notation for the 1st derivative. Substitute in the 1st derivative equation:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]$

Simplifying it, we have:

$\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]$

We can rewrite the 2nd derivative using exponential rules:

$\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}$

To evaluate the 2nd derivative at x = 2, simply substitute in x = 2 and the value f(2) = 2 into it:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}$

When we evaluate this using order of operations, we should obtain our answer:

$\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = 219$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

5 0

2 years ago