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larisa [96]
2 years ago
7

PLS HELP ASAP PLSSS IT DETECTS IF ITS RIGHT OR WRONG HELPPP

Mathematics
1 answer:
pickupchik [31]2 years ago
4 0

Answer:

Step-by-step explanation:

bottom one because x is the variable being changed

meaing that the "per months" is going to always be your x

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3+(4+5) = ( 3+4 ) +5.THIS IS AN EXAMPLE OF THE ---------OF WHOLE
navik [9.2K]

Answer:

D.

Step-by-step explanation:

The problem is clearly addition, so we can rule out B and C. Associative property is the movement of the parentheses, whereas commutative property is the movement or reorganization of the numbers. The parentheses moved in this problem, so it is associative. I suggest reviewing these properites, they come in handy later and are good to know :)

8 0
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What value does the 8 represent in the number 865,202 ?
Lina20 [59]

The value of 8 is 800,000.

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Please answer the image below
nikklg [1K]

Answer:

<h2>301.593</h2><h2 />

Step-by-step explanation:

surface area = 2πrh + 2πr²

where r = 4 km radius

           h = 8 km

<u>plugin values into the formula</u>

surface area = 2πrh + 2πr²

                     = 2π (4) 8  +  2π (4)²

                     = 201.062  +  100.531

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7 0
2 years ago
Is the relation a function?
Margaret [11]
No; a domain value has two range values.

x = -2 then y = 1 and 2

they would form a vertical line, which tells us that it's not a function
8 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

8 0
2 years ago
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