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3 years ago

What is the ratio of 15 to 21

Mathematics

1 answer:

kifflom [539]3 years ago

7 0

Answer: a ratio is a fraction or comparison of two numbers

so you put 15/21 and simplify giving you 5/7

Step-by-step explanation:

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3 years ago

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3 years ago

A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co

butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

$\displaystyle \vec{v}=\frac{\vec{x}}{t}$

Where $\vec x$ is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

$\displaystyle \vec{x}=\vec{v}.t$

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

$\displaystyle \vec{v_1}=$

$\displaystyle \vec{v_1}=\ km/h$

The displacement of the plane in 2 hours is

$\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)$

$\displaystyle \vec{x_1}=km$

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

$\displaystyle \vec{v_2}=$

$\displaystyle \vec{v_2}=km/h$

The displacement in 1 hour is

$\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)$

$\displaystyle \vec{x_2}=km$

The total displacement is the vector sum of both

$\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+$

$\displaystyle \vec{x_t}=km$

$\displaystyle \vec{x_t}=$

The distance from the airport is the module of the displacement:

$\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}$

$\displaystyle |\vec{x_t}|=894.43\ km$

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3 years ago

In a (blank) , one ratio compares a part to a whole

cluponka [151]

I think it is a rate?

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3 years ago

WILL GIVE BRAINLIEST

Natasha_Volkova [10]

Answer:

see explanation

Step-by-step explanation:

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3 years ago