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andrezito [222]
3 years ago
6

A sea turtle swims at a speed of 27 kilometers per hour. A girl swims 14 decimeters per second. 1 m = 10 dm 1000 m = 1 km How mu

ch faster does the sea turtle swim than the girl in meters per minute?
Mathematics
2 answers:
cupoosta [38]3 years ago
5 0
The answer is 377 I took the test.
insens350 [35]3 years ago
4 0
The sea turtle's speed is
v_{t} = (27 \, \frac{km}{h} )*(1000 \, \frac{m}{km} )*( \frac{1}{60} \, \frac{h}{min} ) = 450 \, \frac{m}{min}

The girl's speed is
v_{g} = (14 \, \frac{dm}{s} )*( \frac{1}{10} \, \frac{m}{dm} )*(60 \, \frac{s}{min} ) = 84 \, \frac{m}{min}

The ratio of the turtle's speed to that of the girl is
\frac{v_{t}}{v_{g}} = \frac{450}{84} =5.357

Answer: 5.36 faster  (nearest hundredth)


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Read 2 more answers
Q(a+1) - 2q(a) if q(x) = x²+3x+4​
siniylev [52]

q(x) =  {x}^{2} + 3x + 4

_________________________________

Step(1)

To find q(a) we just need to put a instead of x in q(x) function.

Let's do it...

q(a) =  {a}^{2} + 3a + 4

Multiply sides by -2 :

- 2q(a) =  - 2( {a}^{2} + 3a + 4)

- 2q(a) =  - 2 {a}^{2} - 6a - 8

_________________________________

Step (2)

To find q(a+1) we just need to put a+1 instead of x in q(x) function.

Let's do it...

q(a + 1) =  ({a + 1})^{2} + 3(a + 1) + 4 \\

q(a + 1) =  {a}^{2} + 2a + 1 + 3a + 3 + 4 \\

q(a + 1) =  {a}^{2} + 5a + 8

_________________________________

Step (3)

q(a + 1) - 2q(a) =

{a}^{2} + 5a + 8 - 2 {a}^{2} - 6a - 8 =  \\

-  {a}^{2} - a  =  - a(a + 1)

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

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