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3 years ago

Which expression is equivalent to ^3√x^5y?a)x5/3b)x5/3y1/3c)x3/5 yd)x3/5 y3

Mathematics

1 answer:

Illusion [34]3 years ago

7 0

Answer:

$(x^5y)^{\frac{1}{3}}= x^{\frac{5}{3}} \times y^{\frac{1}{3}}$

Step-by-step explanation:

We are given our expression as

$\sqrt[3]{x^5y}$

The property of exponent says that

$\sqrt[n]{x} =x^{\frac{1}{n}}$

Hence

$\sqrt[3]{x^5y}= (x^5y)^{\frac{1}{3}}$

Another property says

$(ab)^n=a^n \times b^n$

Hence

$(x^5y)^{\frac{1}{3}}= x^{\frac{5}{3}} \times y^{\frac{1}{3}}$

Hence option B is correct.

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graph the function f (x) = x^2 + 4x -5. on the coordinate plane. (a) What are the x-intercepts? (b) What are the y-intercepts? (

Butoxors [25]

Answer:

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

See attached graph.

Step-by-step explanation:

To graph the function, find the vertex of the function find (-b/2a, f(-b/2a)). Substitute b = 4 and a = 1.

-4/2(1) = -4/2 = -2

f(-2) = (-2)^2 + 4(-2) - 5 = 4 - 8 - 5 = -4 - 5 = -9

Plot the point (-2,-9). Then two points two points on either side like x = -1 and x = -3. Substitute x = -1 and x = -3

f(-1) = (-1)^2 + 4 (-1) - 5 = 1 - 4 - 5 = -8

Plot the point (-1,-8).

f(-3) = (-3)^2 + 4(-3) - 5 = 9 - 12 - 5 = -8

Plot the point (-3,-8).

See the attached graph.

The features of the graph are:

a) (-5,0) and (1,0)

b) (0,-5)

c) minimum

5 0

3 years ago

Y= -4/3x - 5/2 y-intercept: slope:

CaHeK987 [17]

Answer:the answer is Y= -4/3x - 5/2

Step-by-step explanation:

3 0

3 years ago

Can anybody help plzz?? 65 points

Yakvenalex [24]

Answer:

$\frac{dy}{dx} =\frac{-8}{x^2} +2$

$\frac{d^2y}{dx^2} =\frac{16}{x^3}$

Stationary Points: See below.

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Calculus

Derivative Notation dy/dx

Derivative of a Constant equals 0.

Stationary Points are where the derivative is equal to 0.

1st Derivative Test - Tells us if the function f(x) has relative max or mins. Critical Numbers occur when f'(x) = 0 or f'(x) = undef
2nd Derivative Test - Tells us the function f(x)'s concavity behavior. Possible Points of Inflection/Points of Inflection occur when f"(x) = 0 or f"(x) = undef

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

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

$f(x)=\frac{8}{x} +2x$

Step 2: Find 1st Derivative (dy/dx)

Quotient Rule [Basic Power]: $f'(x)=\frac{0(x)-1(8)}{x^2} +2x$
Simplify: $f'(x)=\frac{-8}{x^2} +2x$
Basic Power Rule: $f'(x)=\frac{-8}{x^2} +1 \cdot 2x^{1-1}$
Simplify: $f'(x)=\frac{-8}{x^2} +2$

Step 3: 1st Derivative Test

Set 1st Derivative equal to 0: $0=\frac{-8}{x^2} +2$
Subtract 2 on both sides: $-2=\frac{-8}{x^2}$
Multiply x² on both sides: $-2x^2=-8$
Divide -2 on both sides: $x^2=4$
Square root both sides: $x= \pm 2$

Our Critical Points (stationary points for rel max/min) are -2 and 2.

Step 4: Find 2nd Derivative (d²y/dx²)

Define: $f'(x)=\frac{-8}{x^2} +2$
Quotient Rule [Basic Power]: $f''(x)=\frac{0(x^2)-2x(-8)}{(x^2)^2} +2$
Simplify: $f''(x)=\frac{16}{x^3} +2$
Basic Power Rule: $f''(x)=\frac{16}{x^3}$

Step 5: 2nd Derivative Test

Set 2nd Derivative equal to 0: $0=\frac{16}{x^3}$
Solve for x: $x = 0$

Our Possible Point of Inflection (stationary points for concavity) is 0.

Step 6: Find coordinates

Plug in the C.N and P.P.I into f(x) to find coordinate points.

x = -2

Substitute: $f(-2)=\frac{8}{-2} +2(-2)$
Divide/Multiply: $f(-2)=-4-4$
Subtract: $f(-2)=-8$

x = 2

Substitute: $f(2)=\frac{8}{2} +2(2)$
Divide/Multiply: $f(2)=4 +4$
Add: $f(2)=8$

x = 0

Substitute: $f(0)=\frac{8}{0} +2(0)$
Evaluate: $f(0)=\text{unde} \text{fined}$

Step 7: Identify Behavior

See Attachment.

Point (-2, -8) is a relative max because f'(x) changes signs from + to -.

Point (2, 8) is a relative min because f'(x) changes signs from - to +.

When x = 0, there is a concavity change because f"(x) changes signs from - to +.

3 0

3 years ago

What is the area of each piece of paper that Jodi is cutting out

Reptile [31]

The simplest (and most commonly used) area calculations are for squares and rectangles. To find the areaof a rectangle multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself tofind the area.

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3 years ago

Solve -3/4(8k-24)>-12

lisabon 2012 [21]

$-\dfrac{3}{4}(8k-24) > -12\ \ \ \ |\teext{change the signs}\\\\\dfrac{3}{4}(8k-24) < 12\ \ \ \ |\cdot\dfrac{4}{3}\\\\8k-24 < 12\cdot\dfrac{4}{3}\\\\8k-24 < 4\cdot4\\\\8k-24 < 16\ \ \ \ |+24\\\\8k < 40\ \ \ \ |:8\\\\k < 5\to k\in(-\infty;\ 5)$

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3 years ago

Which expression is equivalent to ^3√x^5y?a)x5/3b)x5/3y1/3c)x3/5 yd)x3/5 y3​

Which expression is equivalent to ^3√x^5y?a)x5/3b)x5/3y1/3c)x3/5 yd)x3/5 y3