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krek1111 [17]
3 years ago
9

I(1, 1) and J (-3,-3) Find the midpoint of the line segment

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
4 0
The Answer is (-1,-1)
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The center of a circle and a point on the circle are given. Write the equation of the circle in standard form.
pogonyaev

Answer:

A.Because ,center is (3,2)and point is (-2-3) .

and answer is (x-3)2+(y+2)2=52.

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3 years ago
Si el velocimetro de un auto marca un 25% mas que la verdadera velocidad y en este momento marca 100km/h entonces la velocidad v
GenaCL600 [577]

Answer:

True speed of the car is 87 km/h.

Step-by-step explanation:

Speed = 100 km/h

hike in speed = 25 %

Let the true speed is v.

v + 15% of v = 100 \\\\v + 0.15 v = 100 \\\\1.5 v = 100\\\\ v = 87  km/h

6 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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3 years ago
an architect has an annual income of $85,750. the income rax the professional has to pay is 6%. What is the amount of income tax
viktelen [127]

Answer:

85,750 x 6% = $5,145.00

Step-by-step explanation:

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