Whats an expression for the volume?

What is the approximate solution of this system of equations?

Effectus [21]

Hmm, I'd say it's the first option.

3 0

4 years ago

20 POINTS AND BRAINLIEST

emmainna [20.7K]

To solve this equation we must first plug in 5 for x.

$\frac{1}{3}$ (5 - 4x)
↓
$\frac{1}{3}$ (5 - 4 · 5)

The next step would be to multiply 4 by 5 to get rid of the parenthesis within our main parenthesis.

$\frac{1}{3}$ (5 - 4 · 5)

$\frac{1}{3}$ (5 - 20)

The next step would be to subtract both of the numbers within the parenthesis.

$\frac{1}{3}$ (-15)

Now we must multiply.

- $\frac{1}{3}$ · 15 = - $\frac{1×15}{3}$

- $\frac{15}{3}$

Now we simplify our answer to get our final answer.

-5

8 0

3 years ago

Geometry help! Please prove the following is a square with full explanation, I'm really lost.

Hoochie [10]

We have to show that this is a square, meaning all the sides are perpendicular and the same distance.

A(3,4), B(2,-2), C(-4,-1), D(-3,5)

For each side we calculate the squared distance and the slope. We don't have to bother to take the square root. We'll subtract the points first because that difference, called the direction vector, goes into both the calculation of the slope and of the distance.

AB=B-A=(2 - 3, -2 - 4) = (-1, -6). slope=-6/-1=6 AB²=(-1)²+(-6)²=37

BC=C-B=(-6, 1). slope=1/-6=-1/6. BC²=(-6)²+1²

CD=D-C=(1, 6). slope=6/1=6 CD²=1²+6²=37

DA=A-D=(6,-1) slope=-1/6 DA²=37

We see the negative reciprocal slopes, alternating between 6 and -1/6, indicating perpendicular sides. We see they all have squared distance 37. Equal perpendicular sides proves its a square.

If we do these problems enough we learn: There's no point taking the square root. In fact, there's really no point computing the slopes and the squared distances; we can see it's a square from the pattern of the direction vectors of the sides: (-1, -6) (-6, 1) (1, 6) (6,-1)

5 0

4 years ago

The curve

kherson [118]

Answer:

Point N(4, 1)

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Coordinates (x, y)
Functions
Function Notation
Terms/Coefficients
Anything to the 0th power is 1
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

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:

Step 1: Define

$\displaystyle y = \sqrt{x - 3}$

$\displaystyle y' = \frac{1}{2}$

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y = (x - 3)^{\frac{1}{2}}$
Chain Rule: $\displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)$
Simplify: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1$
Multiply: $\displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}$
[Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}$
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x - 3}}$

Step 3: Solve

Find coordinates

x-coordinate

Substitute in y' [Derivative]: $\displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply 2 on both sides: $\displaystyle 1 = \frac{1}{\sqrt{x - 3}}$
[Multiplication Property of Equality] Multiply √(x - 3) on both sides: $\displaystyle \sqrt{x - 3} = 1$
[Equality Property] Square both sides: $\displaystyle x - 3 = 1$
[Addition Property of Equality] Add 3 on both sides: $\displaystyle x = 4$