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denis-greek [22]
3 years ago
15

the graph of a sine function is shown. select two points on the midline of the function that are separated by a distance of one

period

Mathematics
1 answer:
lidiya [134]3 years ago
6 0

Answer:

(0,3) and (π,3)

Step-by-step explanation:

The period of a sine graph is the magnitude of the interval over which the graph completes a full cycle.

From the graph, the sine function completed one full cycle in [0,π]

The period is π.

The midline us y=3.

Two points on the midline that are separated by a distance of π are (0,3) and (π,3)

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each morning a gardener uses 24 1/2 gallons of water from a metal reservoir.each afternoon,she adds 15 1/4 gallons of water to t
Alika [10]
Every day, the volume increases by 24 1/2+15 1/4=39 3/4 gallons. So after 5 days, the volume increases by 39 3/4 * 5 = 795/4= 198.75 gallons
3 0
3 years ago
A cannon ball shoots out from a cannon that is 28 inches long at 2700 ft s. What is the
Reika [66]

Answer:

a = 1565217.39 ft / s ^ 2

t = 0.001725 seconds

Step-by-step explanation:

The first thing is to use the same system of units therefore we will pass the 28 inches to feet, like this:

28 in * (1 ft / 12 in) = 2.33 ft

Now yes, we can continue, we have the following data:

vi = 0

vf = 2700 ft / s

the equations in this case are as follows:

vf = vi + a * t

vf = a * t

rearranging for a

a = vf / t (1)

now with the position equation we know that:

x = vi * t + (a * t ^ 2) / 2

x = (a * t ^ 2) / 2 (2)

now replacing (1) in (2), we are left with:

x = (vf / t) * (t ^ 2) / 2

knowing that x would be 2.33 ft, which is when the cannonball exits the cannon.

2.33 = 2700 * t / 2

t = 2.33 * 2/2700 = 0.001725 seconds.

and now replace in (1)

a = vf / t = 2700 / 0.001725 = 1565217.39 ft / s ^ 2

3 0
3 years ago
Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 450 newtons stretches a sprin
vlada-n [284]

Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

F =450N

x_1 = 30cm

x_2 = 60cm

Required

The work done

First, we calculate the spring constant (k)

F = kx_1

450N = k *30cm

k = \frac{450N}{30cm}

k =15N/cm

So:

F = kx_1

F(x) = 15x

The work done using Hooke's law is:

W =\int\limits^a_b {F(x)} \, dx

This gives:

W =\int\limits^{60}_{30} {15x} \, dx

Rewrite as:

W =15\int\limits^{60}_{30} {x} \, dx

Integrate

W =15 \frac{x^2}{2}|\limits^{60}_{30}

This gives:

W =15 *\frac{60^2 - 30^2}{2}

W =15 *\frac{2700}{2}

W =15 *1350

W =20250N-cm

Convert to Nm

W =\frac{20250Nm}{100}

W =202.50Nm

7 0
3 years ago
3 folders cost $2.91.<br> Which equation would help determine the cost of 2 folders?
-Dominant- [34]

Answer:

194

Step-by-step explanation:

2/3X=291

x=194

8 0
2 years ago
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