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jarptica [38.1K]
3 years ago
12

Please help me right now.

Mathematics
1 answer:
allochka39001 [22]3 years ago
5 0

X(3, -1) → x = 3, y = -1

x - no change

y:

calculate the distance between X and y = 1

1 - (-1) = 1 + 1 = 2

calculate the value of y in coordinates of X':

1 + 2 = 3

Answer: (3, 3)

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vfiekz [6]

Answer:

a polynomial with degree coefficient real number can be a negative number

i guessed i am not sure

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Suppose that F(x) = x^3 and G(X) = -3x^3. Which statement best compares the
Alik [6]

Answer:

D. The graph of G(x) is the graph of F(x) flipped over the y-axis and

stretched vertically.

Step-by-step explanation:

since there is a negative, there must be a reflection. the reflection is over the y-axis because it is outside the x-value (-F(x) not F(-x))

the same goes for the vertical stretch. we know it is a vertical stretch because the 3 is outside the x-value (3F(x) not F(3x))

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Question 8 Find the unit vector in the direction of (2,-3). Write your answer in component form. Do not approximate any numbers
slamgirl [31]

Answer:

The unit vector in component form is \hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right) or \hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j.

Step-by-step explanation:

Let be \vec u = (2,-3), its unit vector is determined by following expression:

\hat {u} = \frac{\vec u}{\|\vec u \|}

Where \|\vec u \| is the norm of \vec u, which is found by Pythagorean Theorem:

\|\vec u\|=\sqrt{2^{2}+(-3)^{2}}

\|\vec u\| = \sqrt{13}

Then, the unit vector is:

\hat{u} = \frac{1}{\sqrt{13}} \cdot (2,-3)

\hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right)

The unit vector in component form is \hat{u} = \left(\frac{2}{\sqrt{13} },-\frac{3}{\sqrt{13}}  \right) or \hat{u} = \frac{2}{\sqrt{13}}\,i-\frac{3}{13}\,j.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

23-162=

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