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Pachacha [2.7K]
3 years ago
5

PLEASE HELP!

Mathematics
1 answer:
zubka84 [21]3 years ago
3 0

Answer:

A, B, D, and E

Step-by-step explanation:

Because these two side lengths are equal, the only restriction on the length of the third side is that it is smaller than the combined length of the other two sides. Otherwise, the triangle would not be able to be formed. Therefore, the third side length must be under 12. This means that only answer choices A, B, D, and E are possible. Hope this helps!

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An architect plans to buy 5 stone spheres and 3 stone cylinders. For the same​ amount, she can buy 2 stone spheres and 6 stone c
Vedmedyk [2.9K]

The cost of each stone sphere is $ 36.81

<em><u>Solution:</u></em>

Given that,

An architect plans to buy 5 stone spheres and 3 stone cylinders

For the same​ amount, she can buy 2 stone spheres and 6 stone cylinders

Let "x" be that same amount

Let "a" be the cost of each stone sphere

Cost of each stone cylinder = $ 36.81

Therefore,

x = 5 stone spheres and 3 stone cylinders

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5a + 3(36.81) = 2a + 6(36.81)\\\\5a + 110.43 = 2a + 220.86\\\\5a - 2a = 220.86 - 110.43\\\\3a = 110.43\\\\a = 36.81

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3 years ago
Which property is shown in the problem below?
Studentka2010 [4]

Answer:

Step-by-step explanation:

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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
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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.

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