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zheka24 [161]
3 years ago
11

What should be measured to calculate the average rate at which the water in a cup cools?

Mathematics
1 answer:
tekilochka [14]3 years ago
5 0
The temperature of the water should be measured.
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Is an integer that is the square root of an integer ex 4 which can be written 2times 2
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Julie drew a picture of her rectangular garden,. In her picture,the length of her garden is 7/8 inches and the width is 3/4 inch
stepan [7]

Answer:

21/32 square inches.


Step-by-step explanation:

The area of a rectangle = length x width

The area of her picture = length x widht

= 7/8 x 3/4

= 21/32

The area of her picture is 21/32 square inches.

Yes, I agree the area of the picture is 21/32 square inches.

The area must be in square inches. So she need to specify it.

Thank you.

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3 years ago
Which graph represents the solution set for the compound inequality below?
taurus [48]
X-10>7 is the correct andwer
6 0
3 years ago
We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance
MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

h(t) = -16\cdot t^{2}+256\cdot t+1680 (1)

Where:

h - Height above the ground, measured in feet.

t - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680

h(3\,s) = 2304\,ft

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at t = 5\,s and t = 7\,s:

h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680

h(5\,s) = 2560\,s

h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680

h(7\,s) = 2688\,s

We notice that the small object ascends in the given interval.

\Delta h = h(7\,s)-h(5\,s)

\Delta h = 128\,ft

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that h = 2640\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t-960=0 (2)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 10\,s and t_{2}= 6\,s

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that h = 0\,ft, then (1) is reduced into this second-order polynomial:

-16\cdot t^{2}+256\cdot t +1680 = 0 (3)

All roots of the resulting equation come from the Quadratic Formula:

t_{1} = 21\,s and t_{2} = -5\,s

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0
3 years ago
Convert the measurement. Help me ASAP please
sdas [7]
22.86 cm is the answer
5 0
3 years ago
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