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erma4kov [3.2K]
1 year ago
15

An icicle drips at a rate that can be represented by the function f(x) = −x2 + 11x − 18, where 0 ≤ x ≤ 10 and x is the number of

hours after the sun has risen. When f(x) is a negative number, the icicle is not dripping. Determine the values when the icicle starts and stops dripping.
Mathematics
1 answer:
FrozenT [24]1 year ago
4 0

Calculate f(x) for all numbers between 0 and 10:

f(0) = −0^2 + 11(0) − 18 = -18

f(1) = −1^2 + 11(1) − 18 = -8

f(2) = −2^2 + 11(2) − 18 = 0

f(3) = −3^2 + 11(3) − 18 = 6

f(4) = −4^2 + 11(4) − 18 = 10

f(5) = −5^2 + 11(5) − 18 = 12

f(6) = −6^2 + 11(6) − 18 = 12

f(7) = −7^2 + 11(7) − 18 = 10

f(8) = −8^2 + 11(8) − 18 = 6

f(9) = −9^2 + 11(9) − 18 = 0

f(10) = −10^2 + 11(10) − 18 = -8

When the value is negative it doesn't drip, so it starts and stops dripping when the values = 0

Which is f(2) and f(9), the values are 2 and 9.

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If r is the midpoint of qs rs=2x-4, st= 4x-1 and rt = 8x-43 find qs
sammy [17]

Answer:

QS=68\ units

Step-by-step explanation:

step 1

Find the value of x

we know that

r is the midpoint of qs

so

QR=RS

QS=QR+RS------> QS=2RS -----> equation A

RT=RS+ST ----> equation B

see the attached figure to better understand the problem

Substitute the given values in the equation B and solve for x

8x-43=(2x-4)+(4x-1)

8x-43=6x-5

8x-6x=43-5

2x=38

x=19

step 2

Find the value of RS

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substitute the value of x

RS=2(19)-4

RS=34\ units

step 3

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Remember equation A

QS=2RS

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QS=2(34)=68\ units

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2 years ago
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anastassius [24]

Answer:

d. Set A and Set B

Step-by-step explanation:

We have been given three sets:

Set A: (5, 2), (4, 3), (3, 4), (2, 5)

Set B: (-1, -6), (0, 2), (1, 2), (3, 6)

Set C: (2, 1), (4, 2), (2, 3), (8, 4)

Now we need to state about which of the following sets of ordered pairs represent a function.

We know that a function can't have repeated values in domain that is x-value can't repeat.

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Then set C is not a function.

Hence correct choice is:

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Clockwise moments = Anticlockwise moments
Call the force you apply horizontally to the top left side of the square N.
W = 510 acting from the centre as the mass is distributed evenly.

Take moments about a pivot, in this case the bottom right corner is the pivot.
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