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pav-90 [236]
3 years ago
9

Mr. Mole left his burrow and started digging his way down at a constant rate.

Mathematics
2 answers:
Iteru [2.4K]3 years ago
6 0

Constant rate means it is linear so we don't have to worry about fluctuation.

change in y/ change in x=-18-(-25.2)/5-8=2.4 meters per second

hammer [34]3 years ago
5 0

We have been given that Mr. Mole left his burrow and started digging his way down at a constant rate. The table compares Mr. Mole altitude relative to the ground (in meters) and the time since he left (in minutes).

Mr. Mole's speed will be equal to slope of line passing through the given points.

m=\frac{y_2-y_1}{x_2-x_1}

Let point (5,-18)=(x_1,y_1) and point (8,-25.2)=(x_2,y_2).

Upon substituting these values in slope formula, we will get:

m=\frac{-25.2-(-18)}{8-5}

m=\frac{-25.2+18}{3}

m=\frac{-7.2}{3}

m=-2.4  

Mr. Mole's altitude is getting more and more negative as he digging down.

Since speed cannot be negative, therefore, Mr. Mole's speed is 2.4 meters per minute.

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Solve –19 = x – 4. Question 9 options: A) x = –15 B) x = –23 C) x = –76 D) x = 25
AveGali [126]

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A.) x = -15

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-19+4 = x

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4 0
2 years ago
Help if you can do this please
olga nikolaevna [1]

Answer:

Step-by-step explanation:

a. Since the parabola is compressed by a factor of 1/3 we can state:

  • a parabola is written this way : y=(x-h)²+k
  • h stands for the translation to the left ⇒ 2*3=6
  • k for the units down  ⇒4*3=12

So the equation is : y=(x-6)²+12

b.Here the parabola is stretched by a factor of 2 so we must multiply by 1/2

  • We khow that a parabola is written this way : y=(x-h)²+k
  • (h,k) are the coordinates of the vertex
  • the maximum value is  7*0.5=3.5
  • we khow tha the derivative of a quadratic function is null in the maximum value
  • so let's derivate (x-h)²+k= x²+h²-2xh+k
  • f'(x)= 2x-2h    h is 1 since the axe of simmetry is x=1
  • f'(x)=2x-2 ⇒2x-2=0⇒ x= 1
  • Now we khow that 1 is the point where the derivative is null
  • f(1)=3.5
  • 3.5=(x-1)²+k
  • 3.5= (1-1)²+k⇒ k=3.5

So the equation is : y=(x-1)²+3.5

7.

the maximum height is where the derivative equals 0

  • h= -5.25(t-4)²+86
  • h= -5.25(t²-8t+16)+86
  • h=-5.25t²+42t-84+86
  • h=-5.25t²+42t+2

Let's derivate it :

  • f(x)= -10.5t+42
  • -10.5t+42=0
  • 42=10.5t
  • t= 42/10.5=4

When the height was at max t=4s

  • h(max)= -5.25(4-4)²+86 = 86 m

h was 86m

8 0
3 years ago
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