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Thepotemich [5.8K]
3 years ago
9

This last one I need help on too

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
3 0

Answer:

x=\frac{3}{4}+i\frac{\sqrt{7}}{4},\:x=\frac{3}{4}-i\frac{\sqrt{7}}{4}

Step-by-step explanation:

simplify \frac{-3}{x-2} by putting the negative sign on the outside. \frac{2x}{x-1}-\frac{2x-5}{x^2-3x+2}=-\frac{3}{x-2}

find the LCM of the denominators. It is (x-1)(x-2). Multiply by the LCM:

\frac{2x}{x-1}\left(x-1\right)\left(x-2\right)-\frac{2x-5}{x^2-3x+2}\left(x-1\right)\left(x-2\right)=-\frac{3}{x-2}\left(x-1\right)\left(x-2\right)

Simplify:

2x\left(x-2\right)-\left(2x-5\right)=-3\left(x-1\right)

solve: x=\frac{3}{4}+i\frac{\sqrt{7}}{4},\:x=\frac{3}{4}-i\frac{\sqrt{7}}{4}

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Step-by-step explanation:

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to shift down by 4, subtract 4 outside of the absolute value.

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5 0
3 years ago
An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

  is the time

18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

- When the rocket hits the ground h(x) = 0

∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

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Answer:

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Let's call the second studio, yoga studio B.

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So, for 12 classes:

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Yoga Studio B: y=12.5(12)+25, y=175

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Yoga Studio B: y=12.5(10)+25, y=150.

These two numbers are not equal, so Gigi is wrong.

Let me know if this helps!

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