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Nostrana [21]
3 years ago
14

Please help me please !!!

Mathematics
1 answer:
Anon25 [30]3 years ago
4 0

3rd - 25 is the 3rd term in the sequence

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Order each number greatest to least 1/2, -2, square root of 5, -7/4, 2.4<br>PLEASE HELP ME
Luden [163]

Answer:

I think its

2.4, square root of 5, 1/2, -7/4, -2

hope this helps :D

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3 years ago
Which equation is satisfied by all three of the plotted points?
Elan Coil [88]

Answer:

Use desmos graphing calculator plug in your equations and see which one has all points.You can use desmos for a bunch of stuff too

Step-by-step explanation:its C though

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8 0
3 years ago
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The answer is c. a rotation and translation
3 0
3 years ago
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Which measurements could not represent the side lengths of a right triangle?
victus00 [196]

Answer: C

Step-by-step explanation: a^2 + b^2 = c^2

A. 6^2 + 8^2 = 10^2

B. 12^2 + 35^2 = 37^2

C. 4^2 + 6^2 does not equal 10^2.....(16 + 36 = 52   10^2 = 100)

D. 10^2 + 24^2 = 26^2

5 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
2 years ago
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