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kompoz [17]
3 years ago
6

The slope of the graph is equal to

Mathematics
1 answer:
elena-s [515]3 years ago
8 0

Answer:

what slope and graph? sorry!

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from a group of 10 men and 8 women, 5 people are to be selected for a committee so that at least 4 men are on the committee. how
yaroslaw [1]
Only 1 way and that is to have 4 men and 1 woman
4 0
3 years ago
Read 2 more answers
help please! I must solve using descartes law of signs or rational root theorem. show all steps please thanks!​
wel

Answer:

x=2 or x=3

Step-by-step explanation:

x^4 - 5x^3 + 7x^2 - 5x + 6 = 0

Step 1=Factor left side of equation.

(x-2) (x-3) (x^2 + 1) = 0

Step 2=Set factors equal to 0.

x-2 = 0 or x-3 = 0 or x^2 + 1 = 0.

Answer:x=2 or x=3

4 0
3 years ago
Help me with these 2 questions ;(
vichka [17]

Answer:

1.) A' = (4,-4)

   B' = (2,-4)

   C' = (2,-2)

   D' = (4,-2)

2.) A' = (-1,5)

    B' = (-2,1)

    C' = (0,1)

Hope this helps!

And please mark as brainliest!

4 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
Help me pleas. Math is not my thing it read.
crimeas [40]

Answer:

74900

Step-by-step explanation:

just multiply the two numbers.

3 0
3 years ago
Read 2 more answers
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