1. how does the graph of a quadratic function in the form y=a(x−p)2+q change when the value of p is increased by 2, the value of q is decreased by 3, and the sign of a is changed to its opposite?
a) the graph is reflected in the x-axis, and translated 2 units right and 3 units up.
b) the graph is reflected in the y-axis, and translated 2 units right and 3 units down.
c) the graph is reflected in the x-axis, and translated 3 units left and 2 units up.
d) the graph is reflected in the x-axis, and translated 2 units right and 3 units down.
2. on average, jesse can bike five times as fast as he can run. to travel 25 km, he needs 2 h more if he is running than if he is biking. what is his average running speed? round your answer to the nearest whole kilometer per hour.
a) 10km/h
b) 5km/h
c) 20km/h
d) 2km/h
3. the half-life of an isotope is 5.6 days. if there is originally 150g of
the isotope, how long will it be before only 10g remain?
a) 4.3 days
b) 21.9 days
c)11.2 days
d)25.0 days

1. the graph is reflected in the x-axis, and translated 2 units right and 3 units down ⇒ answer D

2. His average running speed is 10 km/h ⇒ answer A

3. There will be 21.9 days before only 10 g remain ⇒ answer B

Step-by-step explanation:

* Lets solve the problems

1.

- The quadratic function in the form y = a(x - p)² + q represented by

  upward parabola with minimum vertex (p , q) if a is a positive value

∵ The sign of a is changed to opposite

∴ The parabola will opened downward with maximum vertex

- That means the parabola reflected across the x-axis

∵ The coordinates of the vertex are (p , q)

∵ p is added by 2 and q is decreased by 3

∴ The new vertex is (p + 2 , q - 3)

- That means the parabola will translated 2 units

  to the right and 3 units down

∴ The graph of the quadratic function y = a(x - p)² + q is reflected

   in the x-axis, and translated 2 units right and 3 units down.

2.

- On average, Jesse can bike five times as fast as he can run

* Let his average speed of run is x

∵ His average speed of run is x

∴ His average speed of bike is 5x

- To travel 25 km, he needs 2 h more if he is running than if he

  is biking

∵ The time = distance/speed

∵ The distance is 25 km

∵ His running speed is x

∴ His time of running = 25/x

∵ His biking speed is 5

∴ His time of biking = 25/5x ⇒ divide up and down by 5 to simplify

∴ His time of biking = 5/x

∵ He needs 2 h more if he is running than if he is biking

- That means the time of running = 2 + the time of biking

∴ 25/x = 2 + 5/x

- Multiply all terms by x

∴ 25 = 2x + 5

- Subtract 5 from both sides

∴ 20 = 2x ⇒ divide the two sides by 2

∴ x = 10

∵ x is his average running speed

∴ His average running speed is 10 km/h

3.

- The half-life of an isotope is 5.6 days

∴ The time of every half-life is 5.6

- There is originally 150 g of  the isotope

* Consider that its wight decreased to half every 5.6 days

∴ W = a(b)^t , where a is the original value and b is the factor of

  decreasing and t is the number of the 5.6 days

∵ a = 150 , W = 10 , b = 0.5

∴ 10 = 150(0.5)^t

- divide both sides by 150

∴ 10/150 = (0.5)^t ⇒ simplify it

∴ 1/15 = (0.5)^t

- Insert ㏑ for both sides

∴ ㏑(1/15) = ㏑(0.5)^t

- Remember ㏑(a)^m = m ㏑(a)

∴ ㏑(1/15) = t㏑(0.5)

- Divide both sides by ㏑(0.5)

∴ t = ㏑(1/15) ÷ ㏑(0.5) = 3.90689

∵ t is the number of the 5.6 days

- To find the number of the days multiply t by 5.6

∴ The number of the days = 3.90689 × 5.6 = 21.87858 ≅ 21.9

∴ There will be 21.9 days before only 10 g remain

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step-by-step explanation:

a

b

d

step-by-step explanation: