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ZanzabumX [31]
4 years ago
13

Please help.. I'm really confused tbh

Mathematics
2 answers:
zzz [600]4 years ago
5 0

X axis is the horizontal axis and y is the vertical.

On a graph, quandrant 1 is the upper right side which requires both x and y to be positive.

Quadrant 2 is upper left, which is negative x and positive y.

Quandrant 3 is lower left, which is negative x and y.

Quadrant 4 is lower right, which is positive x and negative y.

X< 0 is negative and y > 0 is positive.

This matches quadrant 2

The answer is B.

alisha [4.7K]4 years ago
4 0
I think ur answer will be C
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8 0
3 years ago
Tim simplified the difference 1/2p-(1/4p-4) as 3/4p-4. Did he find the correct difference? Explain.
Doss [256]

Answer:

<em>TIM'S ANSWER IS WRONG.</em>

The correct simplification is =\frac{p}{4}+4.

Step-by-step explanation:

Given the difference expression

\frac{1}{2}p-\left(\frac{1}{4}p-4\right)

Tim's answer : \frac{3}{4}p-4

<em />

<em>TIM'S ANSWER IS WRONG. </em>

Let us correctly simplify the difference expression

As

=\frac{1}{2}p-\left(\frac{1}{4}p-4\right)

=\frac{p}{2}-\left(\frac{p}{4}-4\right)

=\frac{p}{2}-\frac{p}{4}+4

Combine Like Terms:

=(\frac{p}{2}-\frac{p}{4})+(4)

=\frac{p}{4}+4          ∵  \frac{p}{2}-\frac{p}{4}=\frac{p}{4}

Hence, the correct simplification is =\frac{p}{4}+4.

<em>Keywords: operation, difference</em>

<em> Learn more operations from brainly.com/question/768264</em>

<em> #learnwithBrainly</em>

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3 years ago
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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
4 years ago
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