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Step2247 [10]
3 years ago
10

A cyclist rides her bike at a speed of 10 miles per hour. What is this speed in kilometers per hour? How many kilometers will th

e cyclist travel in 3 hours? In your computations, assume that 1 mile is equal to 1.6 kilometers. Do not round your answers.
Mathematics
2 answers:
myrzilka [38]3 years ago
7 0

Answer:

16 kilometers per hour

48 kilometers

Step-by-step explanation:

10m/ph -> km/hr

since 1 mile is equal to 1.6 kilometers, we use this to convert.

10miles x 1.6 kilometers

10 x 1.6 = 16

so, the answer is 16 kilometers per hour

To find out how many kilometers they travel in three hours, we multiply our last answer times three.

16 x 3 = 48

48 kilometers

This is because the cyclist is traveling at 16 kilometers per hour so we multiply it by the three hours in the question.

ddd [48]3 years ago
4 0

Answer: 48.2803

Step-by-step explanation: well in 10 km, it's 16.09 so that times 3 hrs/30=48.2803

Since it's 10 miles per (hour) then every hour she would travel 10 miles so by the 3rd hour, she'd have traveled 48.28 kilometers

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3 years ago
This question is realy hard so helpp
LUCKY_DIMON [66]

Answer:

Algebrically the result can be written as -

9 {x}^{2}  + 24x + 16

Step-by-step explanation:

Let the number be 'x'.

  1. When multiplied by 3 , algebrically it is written as "3x".
  2. When 4 is added to 3x , algebrically it is written as "3x + 4".
  3. When (3x + 4) is squared completely , algebrically it is written as - {(3x + 4)}^{2}  = 9 {x}^{2}  + 24x + 16
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2 years ago
Factor this polynomial -3x^2-16x-5 ​
dlinn [17]
The answer is -(3x+1)(x+5)
6 0
3 years ago
Please help i dont understand what its asking
Arisa [49]

Multiply both the numerator and denominator by 2+\sqrt3, which is called the "conjugate" of 2-\sqrt3:

\dfrac{5+\sqrt3}{2-\sqrt3}\cdot\dfrac{2+\sqrt3}{2+\sqrt3}

Why do we pick this number? Recall the difference of squares factorization:

a^2-b^2=(a-b)(a+b)

Now replace a=2 and b=\sqrt3. Then

(2-\sqrt3)(2+\sqrt3)=2^2-(\sqrt3)^2=4-3=1

So in your fraction, you end up with

\dfrac{5+\sqrt3}{2-\sqrt3}=\dfrac{(5+\sqrt3)(2+\sqrt3)}1=(5+\sqrt3)(2+\sqrt3)

Finally, just expand the product.

(5+\sqrt3)(2+\sqrt3)=10+7\sqrt3+(\sqrt3)^2=\boxed{13+7\sqrt3}

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Which transformation would take Figure A to Figure B?​
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