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masha68 [24]
3 years ago
10

What are the factors of 21

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
6 0
1,3,7, and 21.
:D :D :D :D :D

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Write an equation of the direct variation that includes the point (–10, –17).
Ganezh [65]

Answer: i think the answer is 17 x-10y=0

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A box of cereal states that there are 96 calories in a 3/4 cup serving what is the unit rate for calories per cup? How many cups
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3 0
3 years ago
Find the following quantity. 6% of 124 7.44 0.744 74.4 744
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Read 2 more answers
Which two transformations must be applied to the graph of y = ln(x) to result in the graph of y = –ln(x) + 64?
stiks02 [169]

Answer: A) reflection over the x-axis, plus a vertical translation

Step-by-step explanation:

Ok, when we have a function y = f(x)

> A reflection over the x-axis changes a point (x, y) to a point (x, -y), then for a function (x , y = f(x)) the point will change to (x, -y =- f(x))

then for a funtion g(x), this tranformation can be written as h(x) = -g(x).

> A vertical translation of A units (A positive) up for a function g(x) can be written as: h(x) = g(x) + A.

Then in this case we have:

y = g(x) = ln(x)

and the transformed function is h(x) = -ln(x) + 64

Then we can start with h(x) = g(x)

first do a reflection over the x-axis, and now we have:

h(x) = -g(x) = -ln(x)

And now we can do a vertical translation of 64 units up

h(x) = -g(x) + 64 = -ln(x) + 64

Then the correct option is:

A) reflection over the x-axis, plus a vertical translation

3 0
3 years ago
Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

7 0
3 years ago
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