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Mnenie [13.5K]
3 years ago
5

Can someone help me with this problem?

(18x^2+9x-7)/3x" alt="(18x^2+9x-7)/3x" align="absmiddle" class="latex-formula">

Mathematics
1 answer:
Fantom [35]3 years ago
5 0
I hope this helps you

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usain bolt, a Jamaica retired sprinter and world record holder in the 100 meters, 200 meters and 4 × 100 meters relay, has an av
Novosadov [1.4K]

For this case we have the following conversion of units:

1mi = 1609m

1h = 3600s

Thus, applying the conversion of units we have:

(23.35 \frac{mi}{h}) (\frac{1609}{1}\frac{m}{mi}) (\frac {1}{3600}\frac{h}{s}) = 10.4 \frac{m}{s}

Then, the distance traveled for 5 seconds is given by:

d = (10.4) (5)

d = 52m

Answer:

he has completed 52 meters of the 80-meter race

5 0
3 years ago
Find the length of a line segment CD with endpoint C at (-3,1) and endpoint D at (5,6)
lutik1710 [3]
I believe your length would be 8, because -3 to get to 0 is 3 then after that you still have 5 so you add them and get 8
8 0
3 years ago
Can someone give me the answers to this or how to find the answers to it?
faltersainse [42]

the discriminant formula is b^2-4ac

so plug the values from each equation into the formula and solve, the result is the value of the discriminant

if the number is negative, there are no real roots/x-int

if it is 0 there is one real root/x-intercepts

if it is positive it has 2 real roots/x-int

and to find the actual solutions you have to plug the values into the quadratic formula

6 0
3 years ago
I NEED THIS ASAP, PLEASE HELP!!!
WINSTONCH [101]

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

7 0
3 years ago
Please help :D <3. Do not understand this one.
Shtirlitz [24]
Sorry for my messy handwriting but this is the answer

8 0
3 years ago
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