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3 years ago

Please help: 12x^4(-5/6x^3y^2) If possible, include steps Thanks.

1 answer:

4 0

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Question
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$12x^4(- \frac{5}{6} x^3y^2)$

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Multiply the constant: 12 x (-5/6)
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$-10x^4( x^3y^2)$

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Multiply the x (we add the power together when we multiply)
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$-10x^7y^2$

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Answer: -10x⁷y²
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4 years ago

What is a situation in which a polynomial model might make sense, and why?

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Answer and Step-by-step explanation:

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3 years ago

**60 POINTS** How to solve step-by-step...:

Katen [24]

sqrt(3x+7)=x-1 One solution was found : x = 6
Radical Equation entered :

√3x+7 = x-1

Step by step solution :Step 1 :Isolate the square root on the left hand side :

Radical already isolated
 √3x+7 = x-1

Step 2 :Eliminate the radical on the left hand side :

Raise both sides to the second power
 (√3x+7)2 = (x-1)2

 After squaring
 3x+7 = x2-2x+1

Step 3 :Solve the quadratic equation :

Rearranged equation
 x2 - 5x -6 = 0

 This equation has two rational roots:
 {x1, x2}={6, -1}

Step 4 :Check that the first solution is correct :

Original equation
 √3x+7 = x-1

 Plug in 6 for x
 √3•(6)+7 = (6)-1

 Simplify
 √25 = 5
 Solution checks !!
 Solution is:
 x = 6

Step 5 :Check that the second solution is correct :

Original equation
 √3x+7 = x-1

 Plug in -1 for x
 √3•(-1)+7 = (-1)-1

 Simplify
 √4 = -2
 Solution does not check
 2 ≠ -2

One solution was found : x = 6

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3 years ago

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A photographer has a print that is 3.5 inches wide and 2.5 inches tall. If the photographer enlarges the print so that it is 7 i

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B. if it is 7 inches wide it will be 6 inches tall

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3 years ago

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A ladder 20 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2

alukav5142 [94]

Answer:

0.17 °/s

Step-by-step explanation:

Since the ladder is leaning against the wall and has a length, L and is at a distance, D from the wall. If θ is the angle between the ladder and the wall, then sinθ = D/L.

We differentiate the above expression with respect to time to have

dsinθ/dt = d(D/L)/dt

cosθdθ/dt = (1/L)dD/dt

dθ/dt = (1/Lcosθ)dD/dt where dD/dt = rate at which the ladder is being pulled away from the wall = 2 ft/s and dθ/dt = rate at which angle between wall and ladder is increasing.

We now find dθ/dt when D = 16 ft, dD/dt = + 2 ft/s, and L = 20 ft

We know from trigonometric ratios, sin²θ + cos²θ = 1. So, cosθ = √(1 - sin²θ) = √[1 - (D/L)²]

dθ/dt = (1/Lcosθ)dD/dt

dθ/dt = (1/L√[1 - (D/L)²])dD/dt

dθ/dt = (1/√[L² - D²])dD/dt

Substituting the values of the variables, we have

dθ/dt = (1/√[20² - 16²]) 2 ft/s

dθ/dt = (1/√[400 - 256]) 2 ft/s

dθ/dt = (1/√144) 2 ft/s

dθ/dt = (1/12) 2 ft/s

dθ/dt = 1/6 °/s

dθ/dt = 0.17 °/s

8 0

3 years ago