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2 years ago

Calculate the side lengths of the given scale.

Mathematics

1 answer:

Amiraneli [1.4K]2 years ago

3 0

Answer:

x: 4metres

y: 8metres

Step-by-step explanation:

using scale 1:4

x:16 ÷ 4

y: 2×4

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Which two values of x satisfy the equation square root 3-2cos x = 2?

Andrei [34K]

Answer: x=2π/3 and x=4π/3

Step-by-step explanation:

The equation we have is $\sqrt{3-2cosx} =2$ . All we have to do is get cosine alone to find the 2 values of x.

$3-2cosx=4\\-2cosx=1\\cosx=-\frac{1}{2}$

Now that we have our cosine left, we can use our unit circle to figure out when does cosx=-1/2. Cosine is the x value of the coordinate.

x=2π/3

x=4π/3

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3 years ago

If x and y are both negative when is x-y positive

Vladimir [108]

X-y is positive when y is further away from 0, compared to x. In other words, when y is a larger negative

For example
x = -2
y = -5

x-y = -2-(-5) = -2+5 = 3

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4 years ago

Find the following differences:

Lyrx [107]

Answer:

1- 0.123

2- 0.0745

3- 0.9706

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3 years ago

What theorem can be used to prove that the two triangles are similar?

kupik [55]

Hey!

Your answer will be:

Angle-Angle (AA) Similarity Postulate!

(Explanation) If two angles of one triangle are congruent to two angles of another, then the triangles must be similar.

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3 years ago

Read 2 more answers

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 $-7.6\:\frac{m}{s}$

7 0

3 years ago