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Papessa [141]
2 years ago
15

Using the numbers 0-6 once, make 2 equivalent ratios

Mathematics
1 answer:
fenix001 [56]2 years ago
8 0

Step-by-step explanation:

As we have to use the number 0-6 once to make 2 equivalent ratios.

so lets solve the problem.

Determining first equivalent ratio

as

  • 5 times 2 = 10

and

  • 32 times 2 = 64

so

  • Together will get 10/64

Therefore, first equivalent ratio

\:\frac{5}{32}=\frac{10}{64}

Determining second equivalent ratio

\frac{13}{65}=\frac{4}{20}

as

\frac{13}{65}=\frac{1}{5}

and

\frac{4}{20}=\frac{1}{5}

Therefore, second equivalent ratio

\frac{13}{65}=\frac{4}{20}

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A road perpendicular to a highway leads to a farmhouse located d miles away. An automobile traveling on this highway passes thro
pshichka [43]

Answer:

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

Step-by-step explanation:

A road is perpendicular to a highway leading to a farmhouse d miles away.

An automobile passes through the point of intersection with a constant speed \frac{dx}{dt} = r mph

Let x be the distance of automobile from the point of intersection and distance between the automobile and farmhouse is 'h' miles.

Then by Pythagoras theorem,

h² = d² + x²

By taking derivative on both the sides of the equation,

(2h)\frac{dh}{dt}=(2x)\frac{dx}{dt}

(h)\frac{dh}{dt}=(x)\frac{dx}{dt}

(h)\frac{dh}{dt}=rx

\frac{dh}{dt}=\frac{rx}{h}

When automobile is 30 miles past the intersection,

For x = 30

\frac{dh}{dt}=\frac{30r}{h}

Since h=\sqrt{d^{2}+(30)^{2}}

Therefore,

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+(30)^{2}}}

\frac{dh}{dt}=\frac{30r}{\sqrt{d^{2}+900}}

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