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maria [59]
2 years ago
5

B is between A and C, if AB=2(x+1), BC=3x+1, and AC=4(x+2), find the value of AC

Mathematics
1 answer:
hoa [83]2 years ago
3 0
2x + 2 + 3x + 1 = 4x + 8

5x + 3 = 4x + 8

5x = 4x - 5

x = -5

therefore

4(-5 + 2)

4(-3)

-12

or

4(-5 + 2)

-20 + 8

-12

either way, it’s the same.

————

answer: AC = -12
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A steamer travels 36 km upstream and 32 km downstream in 6.5 hours. The same steamer travels 4 km upstream and 40 km downstream
rewona [7]

Answer:

The streamer's speed in still water is 90.23 km/h while the stream's speed is 33.33 km/h

Step-by-step explanation:

Let v = streamer's speed in still water and v' = stream's speed. His speed upstream is V = v + v' and his speed downstream is V' = v - v'.

Since he travels 36 km upstream, the time taken is t = 36/V = 36/(v + v').

he travels 32 km downstream, the time taken is t' = 32/V' = 32/(v - v')

The total time is thus t + t' = 36/(v + v') + 32/(v - v')

Since the whole trip takes 6,5 hours,

36/(v + v') + 32/(v - v') = 6.5  (1)

Multiplying each term by (v + v')(v - v'), we have

(v + v')(v - v')36/(v + v') + (v + v')(v - v')32/(v - v') = 6.5(v + v')(v - v')  (1)

(v - v')36 + (v + v')32 = 6.5(v + v')(v - v')  (1)

36v - 36v' + 32v + 32v' = 6.5(v² + v'²)

68v - 4v' = 6.5(v² + v'²)        (2)

Also he travels 4 km upstream, the time taken is t" = 4/V = 4/(v + v').

he travels 40 km downstream, the time taken is t'" = 40/V' = 40/(v - v')

The total time is thus t" + t'" = 4/(v + v') + 40/(v - v')

Since the whole trip takes 180 minutes = 3 hours,

4/(v + v') + 40/(v - v') = 3  (3)

Multiplying each term by (v + v')(v - v'), we have

(v + v')(v - v')4/(v + v') + (v + v')(v - v')40/(v - v') = 3(v + v')(v - v')  (1)

(v - v')4 + (v + v')40 = 3(v + v')(v - v')  (1)

4v - 4v' + 40v + 40v' = 3(v² + v'²)

44v - 36v' = 3(v² + v'²)      (4)

Dividing (2) by (4), we have

(68v - 4v')/(44v - 36v') = 6.5(v² + v'²)/3(v² + v'²)      

(68v - 4v')/(44v - 36v') = 6.5/3

3(68v - 4v') = 6.5(44v - 36v')

204v - 12v' = 286v - 234v'

204v - 286v = 12v' - 234v'

-82v = -222v'

v = -222v'/82

v = 111v'/41

Substituting v into (2), we have

68v - 4v' = 6.5(v² + v'²)      

68(111v'/41) - 4v' = 6.5[(111v'/41)² + v'²]        

[68(111/41) - 4]v' = 6.5[(111/41)² + 1]v'²    

[68(111/41) - 4]v' = 6.5[(111/41)² + 1]v'²

[7548/41 - 4]v' = 6.5[12321/1681 + 1]v'²

[(7548 - 164)/41]v' = 6.5[(12321 + 1681)/1681]v'²

[7384/41]v' = 6.5[14002/1681]v'²

[7384/41]v' = [91013/1681]v'²

[91013/1681]v'² - [7384/41]v' = 0

([91013/1681]v' - [7384/41])v' = 0

⇒ v' = 0 or ([91013/1681]v' - 7384/41) = 0

⇒ v' = 0 or [91013/1681]v' - 7384/41) = 0

⇒ v' = 0 or v' =  7384/41 × 16841/91013

⇒ v' = 0 or v' =  180.097 × 0.185

⇒ v' = 0 or v' =  33.33 km/h

Since v' ≠ 0, v' = 33.33 km/h

Substituting v' into v = 111v'/41 = 111(33.33 km/h)/41 = 3699.63 km/h ÷ 41 = 90.23 km/h

So, the streamer's speed in still water is 90.23 km/h while the stream's speed is 33.33 km/h

5 0
3 years ago
5x 2(x - 3) = -2(x - 1)
natali 33 [55]
0000000000000000000000000000000
4 0
3 years ago
PLEASE HELP!!!
Ksju [112]
ANSWER

a = 1,2,3,6

EXPLANATION

The given equation is

ax = 6

Divide both sides by a

\Rightarrow \: \frac{ax}{a} = \frac{6}{a}

Cancel out common factors on the left hand side,

\Rightarrow \: x = \frac{6}{a}

For this solution to be a whole number, thena should be a factor of
6.

The reason is that, factors of 6 will divide exactly in to 6 without a remainder, making the answer a whole number.

The whole numbers that are factors of 6 are,

1,2,3 \: and \: 6

Note that, the set of whole numbers are,

{0,1,2,3,4,...}
3 0
3 years ago
Read 2 more answers
Jaelynn has 30 times as many stikers as her brother her brother in has 8 stikers how many stikers does jaelynn have
Andre45 [30]

30 × 8 = 240 is the answer

6 0
3 years ago
Simplify the midpoint of QS.
oee [108]

Answer:

[(a + b), c]

Step-by-step explanation:

Midpoint of a segment with extreme ends represented by ordered pairs (x_1,y_1) and (x_2,y_2),

Midpoint = (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

It is given that extreme ends of the segment QS are Q(2b, 2c) and S(2a, 0)

Coordinates of the midpoint of QS will be,

= (\frac{2a+2b}{2},\frac{2c+0}{2} )

= [(a + b), c]

Therefore, ordered pair representing the midpoint of QS will be [(a + b), c].

8 0
3 years ago
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