Please answer due today!

The volume of this rectangular prism is 4x^3 (4x cubed). What does the coefficient 4 mean in terms of the problem?

yKpoI14uk [10]

It means that has four of those cubes inside one prism... it's hard to explain but x^3 is ONE CUBE not prism, YOU HAVE FOUR of them, hope this helpz

5 0

3 years ago

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If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

algol13

Answer:

$\frac{d^2y}{dx^2} = \frac{-4}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate y' terms: $-6x^2=2yy'+y'$
[Algebra] Factor y': $-6x^2=y'(2y+1)$
[Algebra] Isolate y': $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

Step 3: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute y': $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

Step 4: Find Slope at Given Point

[Algebra] Substitute in x and y: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$

6 0

3 years ago

Suppose two acid solutions are mixed. One is 26% acid and the other is 34% acid. Which one of the following concentrations canno

valentina_108 [34]

Answer:

A

Step-by-step explanation:

7 0

3 years ago

Look at the ingredients to make honey biscuits. Work out how much of each ingredient is needed to make: a) 100 honey biscuits. b

Svetllana [295]

Answer:

look bellow

Step-by-step explanation:

100:

25x4=100

50x4=200g

75x4=300g

100x4=400ml

40:

25x1.6=40

50x1.6=80g

75x1.6=120g

100x1.6=160ml

6 0

3 years ago

The speed of sound in air is a linear function of the air temperature. When the air temperature is 10°C, the speed of sound is 3

Sergeu [11.5K]

9514 1404 393

Answer:

process: substitute the given point values into the 2-point form of the equation for a line

Step-by-step explanation:

There are more than a half-dozen different forms of the equation for a line. They are useful for different purposes. One of them is the "two-point form".

Using x as the independent variable, and y as the dependent variable, the equation can be written as ...

y -y1 = (y2 -y1)/(x2 -x1)/(x -x1)

where (x1, y1) and (x2, y2) are the two points.

__

Here, your two points are ...

(t, s) = (10, 337) and (30, 349)

Using s in place of y, and t in place of x, these two points go into the formula like this:

s -337 = (349 -337)/(30 -10)(t -10)

Simplifying the fraction, this is ...

s -337 = (12/20)(t -10)

And writing it as a decimal, we get ...

s -337 = 0.6(t -10)

_____

Additional comments

Adding y1 to both sides of the above form gives you ...

y = (y2 -y1)/(x2 -x1)/(x -x1) +y1

This is the form I usually prefer to use, because it can lead directly to slope-intercept form. For this problem, the form shown above gets you to the answer you're looking for.

__

This "two-point form" is an expansion of the "point-slope form", which is ...

y - k = m(x -h) . . . . . . . line with slope m through point (h, k)

where the equation for slope is ...

m = (y2 -y1)/(x2 -x1)

and (x1, y1) is used instead of (h, k).

5 0

3 years ago

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