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AleksandrR [38]
3 years ago
6

Tell whether the ratios form a proportion 1/3 and 7/21

Mathematics
1 answer:
wolverine [178]3 years ago
3 0
It is proportional ;)
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Describe the transformation from y= x + 4 to y= .5x + 4 Group of answer choices compresses stretches shifts right 4 shifts left
Nina [5.8K]

Answer:

This type of transformation is a horizontal stretch.

<em></em>

Step-by-step explanation:

Given

y = x + 4

y = .5x + 4

Required

Determine the type of transformation

The first function can be expressed as:

f(x) = x + 4

While the second function is:

g(x) = f(0.5x)

Solving f(0.5x), we have to substitute 0.5x for x in f(x) = x + 4

f(0.5x) = 0.5x + 4

So:

The second function is:

g(x) = 0.5x + 4

<em>This type of transformation is a horizontal stretch.</em>

<em></em>

<em>i.e. f(x) stretched to g(x)</em>

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3 years ago
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Aleksandr-060686 [28]

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Step-by-step explanation:

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It would like this for solving n

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(cot^2x - 1)/(csc^2x) = cos2x​
Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

cos(2x)=cos^2x-sin^2x,

cos(2x)=1-2sin^2x, and

cos(2x)=2cos^2x-1

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants.  Rewriting gives you:

\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }

Notice I also wrote the 1 in terms of sin^2(x).

Now we will put the numerator of the bigger fraction over the common denominator:

\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}

And canceling out the sin^2 x leaves us with just

cos^2x-sin^2x which is one of our identities.

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