All categories

2 years ago

How high is the building (ie. from what height was the potato launched) if h=-2(t-3)2+32

Mathematics

1 answer:

Llana [10]2 years ago

5 0

Answer:

Step-by-step explanation:

To find the height from which the potato was launched, you need to set the time to 0 (since that is at the beginning). Plugging this into the equation, you get:

$h=-2(0-3)^2+32$

h=-2(9)+32

h=32-18

h=14

Hope this helps!

You might be interested in

PLEASE HELP ME SO I CAN GRADUATE!!!!!

Aleks04 [339]

Answer:

A: -41/9

Step-by-step explanation:

Substitute the point (4, -5) and slope -1/9 into the equation y = mx + b.

-5 = -1/9(4) + b

-5 = -4/9 + b

Solve for b:

b = -41/9

Our equation: y = -1/9(x) - 41/9

7 0

2 years ago

A standard toilet uses 1.6 gallons per flush. If a toilet is flushed five times per day, how much water will be used in a month

djyliett [7]

Answer:

I believe the answer is 48 gallons.

Step-by-step explanation:

Multiply 1.6 by 30 and you get 48

6 0

2 years ago

Read 2 more answers

10x^3 y^-5 z^-2 if x=3 y=2 and z=5 express your answer as a reduced fraction

sineoko [7]

Answer:

$\large\boxed{\dfrac{27}{80}}$

Step-by-step explanation:

Put x = 3, y = 2 and z = 5 to the given expression $10x^3y^{-5}z^{-2}$ :

$10(3^3)(2^{-5})(5^{-2})\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{10(27)}{(2^5)(5^2)}=\dfrac{270}{(32)(25)}=\dfrac{270}{800}=\dfrac{27}{80}$

7 0

2 years ago

A wind turbine is located at the top of a hill where the wind blows steadily at 12 m/s, and stands 37 m tall. The air then exits

marta [7]

Answer:

Power generated = 4.315 kW

Step-by-step explanation:

Given,

speed of wind when enters into the turbine, V = 12 m/s

speed of wind when exits from the turbine, U = 9 m/s

mass flow rate of the wind, m = 137 kg/s

According to the law of conservation of energy

Energy generated = change in kinetic energy

Since, the air is exiting at same elevation. So, we will consider only kinetic energy.

Energy generated in one second will be given by,

$E\ =\ \dfrac{1}{2}.mV^2-\dfrac{1}{2}.m.U^2$

$=\ \dfrac{1}{2}\times 137\times (12)^2-\dfrac{1}{2}\times 137\times (9)^2$

=4315.5 J

= 4.315 kJ

So, energy is generated in one second = 4.315 kJ

Power generated can be given by,

$P\ =\ \dfrac{\textrm{energy generated}}{time}$

And the energy is generated is already in per second so power generated will be 4.315 kW.

8 0

2 years ago

Why is the adjusted average used in math?

Tatiana [17]

It’s used to correct statistical averages that include obvious imbalances because of outliers in the data set.

3 0

2 years ago