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3 years ago

1.What is the length of the altitude of an equilateral triangle with side lengths of 34 inches?

Mathematics

1 answer:

sweet [91]3 years ago

7 0

Answer:

Step-by-step explanation:

Remark

<u>Question One</u>

It's equilateral. All three sides are equal. The base is 34 as well. So two right angles can be formed on the base by dropping the altitude from the top vertex.

Givens

altitude = h

a = 34/2 = 17

hypotenuse = 34

Formula

a^2 + h^2 = c^2

Solution

17^2 + h^2 = 34^2

289 + h^2 = 1156

h^2 +289 =1156

h^2 = 1156 - 289

h^2 = 867

sqrt(h^2) = sqrt(867)

h = 29.444

==============

<u>Question Two</u>

Givens

b = 10 feet

c = ?

height = h = 16

Formula

c^2 = b^2 + h^2

Solution

c^2 = 10^2 + 16^2

c^2 = 100 + 256

c^2 = 356

sqrt(c^2) = sqrt(356)

c = 18.87

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3 years ago

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with:

frez [133]

Answer:

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$ .

Step-by-step explanation:

This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant $\frac{dx}{dt}$ and we want to find the other rate $\frac{dy}{dt}$ at that instant.

We know the rate of change of x-coordinate and y-coordinate:

$\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}$

We want to find the rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12.

The distance of a point (x, y) and the origin is calculated by:

$s=\sqrt{x^2+y^2}$

We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.

If we apply implicit differentiation in the formula of the distance we get

$s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Substituting the values we know into the above formula

$s=\sqrt{9^2+12^2}=15$

$\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}$

The rate of change of the distance $\frac{ds}{dt}$ when x = 9 and y = 12 is $-7.6\:\frac{m}{s}$

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3 years ago