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sattari [20]
3 years ago
10

If Jana bikes 15 miles in 50 minutes. How can you find the number of miles she bikes in one hour?

Mathematics
1 answer:
adell [148]3 years ago
8 0

Answer:

Therefore,

Jana bikes 18 miles in one hour.

Step-by-step explanation:

Given:

Jana bikes,

15 miles  ---------------> in 50 minutes

To Find:

How many miles  ------------------> in 1 hour = 60 minutes

Solution:

Let 'x' miles Jana Bikes in 1 hour or 60 minutes

As, 1 hour = 60 minutes

The Ratio of Equation will be

\dfrac{15}{x}=\dfrac{50}{60}

Cross Multiplying we get

x=\dfrac{900}{50}=18\ miles

Therefore,

Jana bikes 18 miles in one hour.

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A41.0m guy wire is attached to the top of a 32.8 m antenna and to a point on the ground. How far is the point on the ground from
Setler79 [48]

Answer:

Step-by-step explanation:

A right angle triangle is formed.

The length of the guy wire represents the hypotenuse of the right angle triangle.

The height of the antenna represents the opposite side of the right angle triangle.

The distance, h from base of the antenna to the point on the ground to which the antenna is attached represents the adjacent side of the triangle.

To determine h, we would apply Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side²

Therefore,

41² = 32.8² + h²

1681 = 1075.84 + h²

h² = 1681 - 1075.84 = 605.16

h = √605.16

h = 24.6 m

To determine the angle θ that the wire makes with the ground, we would apply the the cosine trigonometric ratio.

Cos θ = adjacent side/hypotenuse. Therefore,

Cos θ = 24.6/41 = 0.6

θ = Cos^-1(0.6)

θ = 53.1°

7 0
3 years ago
A closed box with a square base is to have a volume of 13 comma 500 cm cubed. The material for the top and bottom of the box cos
andrezito [222]

Answer:

x  =  1,5 cm

h  =  6  cm

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Step-by-step explanation:

Volume of the box is :

V(b)  = 13,5 cm³

Aea of the top is equal to area of the base,

Let call  " x " side of the base then as it is square area is A₁ = x²

Sides areas are 4 each one equal to x * h  (where h is the high of the box)

And volume of the box is   13,5 cm³  = x²*h

Then   h  =  13,5/x²

Side area is :  A₂ =  x* 13,5/x²

A(b)  = A₁  + A₂

Total area of the box as functon of x is:

A(x)  = 2*x²  + 4* 13,5 / x

And finally cost of the box is

C(x)  = 10*2*x²   +  2.50*4*13.5/x

C(x)  = 20*x²  +  135/x

Taking derivatives on both sides of the equation:

C´(x)  =  40*x   -  135*/x²

C´(x)  = 0     ⇒      40*x   -  135*/x² = 0    ⇒  40*x³ = 135

x³  = 3.375

x  = 1,5 cm

And   h  =  13,5/x²     ⇒   h  =  13,5/ (1,5)²

h = 6 cm

C(min)  = 20*x²  + 135/x

C(min)  = 45  +  90

C(min)  = 135 $

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