The cost of a new car is

H(t)=-16(t-1)^2+16 <br> What is the maximum height the ball reaches and how long does it take?

Aleksandr-060686 [28]

Given t = time and h = height

$h(t) = - 16(t - 1)^{2} + 16$

From the equation.

$y = a(x - h)^{2} + k$

h is the axis od symmetry

(h, k) is the vertex.

In the equation, h is 1 and k is 16.

So the maximum value is 16 at t = 1.

The ball reaches maximum at 16 and it takes t = 1 to reach the maximum point.

4 0

2 years ago

The length of a rectangular floor is 4 feet longer than its width w. The area of the floor is 525 ft^2. A) Write a quadratic equ

mina [271]

Answer:

x^21x+25x-525-0

x^21x+25x-525-0xx^2 - 3.7_) +5^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25ensiah193

4 0

3 years ago

Write the digits in the thousands place

Sloan [31]

What digits? is there specific types?

6 0

3 years ago

Read 2 more answers

PLSSSSS HELP I HAVE BEEN STUCK ON THIS FOR THE PAST HOUR!!!!!!!!!!!!!!!!!!!!!!!!!

vladimir1956 [14]

Answer:

$- 6 \sqrt{2}$