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skelet666 [1.2K]
3 years ago
9

What is 1/3 of 100%?

Mathematics
1 answer:
Andrej [43]3 years ago
7 0
33.3333333333333333333333333333333333333(goes on forever)
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Misha wrote the quadratic equation 0=-x2+4x-7 in standard form. what is the value of c in her equation
Masteriza [31]

Answer:

Final answer is c=-7.

Step-by-step explanation:

Given equation is 0=-x^2+4x-7.

Now question says that Misha wrote the quadratic equation 0=-x2+4x-7 in standard form. Now we need to find about what is the value of c in her equation.

We know that standard form of quadratic equation is given by ax^2+bx+c=0.

compare given equation with the standard form, we find that -7 is written in place of +c

so that means +c=-7

or c=-7

Hence final answer is c=-7.

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4 years ago
The change in the price of a certain brand of cereal from 2010 to 2012 is shown in the table. Year Change (in dollars) 2010 +0.3
TEA [102]
The answer is 240 I think
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4 years ago
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What is the value of the function at x=−2?
Nezavi [6.7K]

Answer:

I would think y equals 2

Step-by-step explanation:

If I'm wrong someone tell me

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3 years ago
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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
3 years ago
-4v-12=-12-4v I need so much help!!!!
tankabanditka [31]

Answer:

0 = 0

Step-by-step explanation:

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