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3 years ago

Geometry help pretty please 15 points plus brainliest plus be my friend x3

2 answers:

5 0

1. Is right 3 down 5, because to get from x(0,0) to x'(3,-5), you need I shift to the right 3 and downwards 5.

2. Counterclockwise 180 degrees. This is because 180 degrees is half of a full rotation (360 degrees), meaning the point would be the same no matter which way you rotated 180 degrees.

3. Clockwise 90 degrees. Going counter clock wise 270 degrees is like going around 3/4 of a circle (because 270/360 is 3/4). Just as well, going clock wise 90 degrees is like going 1/4 of the circle, but in the opposite direction. This means they will both land at the same point.

Hope this Helps!

SVEN [57.7K]3 years ago

4 0

1. X(0,0) --> X'(3,-5).
To go from (0,0) to (3,-5), you must go to the right 3 and down 5.

I cannot answer 2 or 3, as rotations are not my strong suit.

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The difference of the squares of two consecutive odd integers is 24. find the integers

marysya [2.9K]

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3 years ago

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Aneli [31]

Answer:

32 divided by 4 = 8,

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2 years ago

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The price of an item yesterday was $160 . Today, the price fell to 104 . Find the percentage decrease.

babymother [125]

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3 years ago

What is the distance between (-3,5) (7,5)

Firlakuza [10]

Answer:

10 units

Step-by-step explanation:

use distance formula: d = square root of the difference of the y-values minus the difference of the x-values

d = $\sqrt{(5-5)^2+(7--3)^2}$

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3 years ago

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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

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3 years ago