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zzz [600]
3 years ago
11

A train travels 315 km in the same time that a car travels 265 km. If the train travels on average 20 km/h faster than the car,

find the average speed of the car and the time taken to travel 265 km.
Mathematics
1 answer:
aev [14]3 years ago
7 0

Answer: 106 km/h,2.5 hr

Step-by-step explanation:

Given

The train travels 315 km in the same time car travels 265 km

The average speed of the train is 20 kmph faster than car

Suppose, speed of the car is v and it travels 265 km in t hours

\therefore \dfrac{315}{v+20}=\dfrac{265}{v}\\\\\Rightarrow 315v=265\left (v+20\right)\\\\\Rightarrow 315v=265v+5300\\\Rightarrow 50v=5300\\\Rightarrow v=106\ km/h

Time taken to travel is

\Rightarrow t=\dfrac{265}{106}}\\\\\Rightarrow t=2.5\ h

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½ (6x + 8) = 2x + 4 + x
bearhunter [10]

Answer:

True

Step-by-step explanation:

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3 years ago
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Find the value of x which satisfies the following equation.<br> log2(x−1)+log2(x+5)=4
weqwewe [10]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \: x = 3

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \:  log_{2}(x - 1)  log_{2}(x + 5)  = 4

\qquad \tt \rightarrow \:  log_{2} \{(x - 1)(x + 5) \} = 4

[ log (x) + log (y) = log (xy) ]

\qquad \tt \rightarrow \: ( x - 1)(x + 5) =  {2}^{4}

\qquad \tt \rightarrow \:  {x}^{2}  + 5x - x - 5 =  16

\qquad \tt \rightarrow \:  {x}^{2}  + 4x - 5 - 16 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 4x -21 = 0

\qquad \tt \rightarrow \:  {x}^{2}  + 7x - 3x - 21 = 0

\qquad \tt \rightarrow \:  x(x + 7) - 3(x + 7) = 0

\qquad \tt \rightarrow \: (x + 7)(x - 3) = 0

\qquad \tt \rightarrow \: x =  - 7 \:  \: or \:  \: x = 3

The only possible value of x is 3, since we can't operate logarithm with a negative integer in it.

\qquad \tt \rightarrow \: x = 3

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

4 0
2 years ago
57 plus what equals 86
Jet001 [13]

Answer:

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Step-by-step explanation:

57+x=86

x=86-57

x=29

3 0
3 years ago
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The deck of a bridge is suspended 275 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of
Ivenika [448]

Answer:

Our equation for the height is:

y(t) = 275 - 16*t^2.

a) To find the average velocity between two times, t1 and t2, (where t2 > t1) the equation is:

AV = \frac{y(t2) - y(t1)}{t2 - t1}

Then:

i) t1 = 4s, t2 = 4s + 0.1s = 4.1s

The average velocity is:

AV = \frac{(275 - 16*4.1^2) - (275 - 16*4^2)}{4.1 - 4} = \frac{16(4^2 - 4.1^2)}{0.1} = -129.6

And the units will be ft/s, so the average speed is:

-129.6 ft/s

The minus sign is because te pebble is falling down.

ii)  t1 = 4s, t2 = 4s + 0.05s = 4.05s

The average velocity is:

AV = \frac{(275 - 16*4.05^2) - (275 - 16*4^2)}{4.05 - 4} = \frac{16(4^2 - 4.05^2)}{0.05} = -128.8

So the average speed is -128.9 ft/s

iii)  t1 = 4s, t2 = 4s + 0.01s = 4.01s

The average speed is:

AV = \frac{(275 - 16*4.01^2) - (275 - 16*4^2)}{4.01 - 4} = \frac{16(4^2 - 4.01^2)}{0.01} = -128.16

The average speed is -128.16 ft/s.

b) The instantaneous velocity of the pebble after 4 seconds can be obtained by looking at the velocity equation, that is the derivative of the height equation.

v(t) = dy(t)/dt.

v(t) = -2*16*t + 0

Then the velocity at t = 4s is:

v(4s) = -32*4 = -128

The instantaneous velocity at t = 4s is -128 ft/s.

3 0
3 years ago
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