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Ivanshal [37]
3 years ago
12

The Black line is the graph of y=c. Which of these equations could represent the blue line?

Mathematics
2 answers:
Nadusha1986 [10]3 years ago
6 0
3rd one is your answer
galben [10]3 years ago
3 0
C. y = 0.5x.

The coefficient of x is the slope of the line. From the picture, you can see that the slope of the blue line is half the slope of the black line.
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-3x - 5(-x + 3) = -25​
Rainbow [258]

Answer:

Step-by-step explanation:

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Where does the section of the bridge meet ground level? Solve 0 = -x2 + 10x - 8
djverab [1.8K]
In can be less mental effort if the equation is multiplied by -1 first.
.. x^2 -10x +8 = 0
.. x^2 -10x +25 +8 -25 = 0
.. (x -5)^2 = 17
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.. x = 5 ±√17

The values of x that satisfy the given equation are
.. x = 5 -√17 ≈ 0.877
.. x = 5 +√17 ≈ 9.123

4 0
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Wes has been tracking the population of a town. The population was 3,460 last year and has been increasing by 1.35 times each ye
AveGali [126]
After x years the population of a town is 3460*[(1.35)^x]

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3 years ago
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PLEASE HELP
postnew [5]

Answer:

Below

Step-by-step explanation:

Substituting the given values:

f(6) = 6(2/3) - 2 =   cube root of 6^2 - 2 = cube root 36 - 2

f(-6)=  (-6)(2/3) - 2 =   cube root of(-6)^2 - 2 = cube root 36 - 2

So This is true,

f(6) =   cube root of 6^2 - 2 = cube root 36 - 2 = 1.3019

2 * f(3) = 2 *  (cube root of 3^2 - 2 )  =  2 * (cube root of 9 - 2) = 0.1602

So False,

4 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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