What happens to the area of a circle when the radius is tripled?

Mathematics

2 answers:

Zolol [24]3 years ago

8 0

The answer should be the circle tripes

patriot [66]3 years ago

6 0

The circle would triple because the radius is equal to its area

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CMS y’all help me on 5

yulyashka [42]

Answer:

the answer is c

Step-by-step explanation:

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3 years ago

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The coordinates for a rhombus are given as (2a, 0), (0, 2b), (–2a, 0), and (0, –2b). Write a plan to prove that the midpoints of

Verizon [17]

The midpoints of the sides are:
M ( (2a+0)/2 , (2b+0)/2 ) = ( a , b )
N ( (-2a+0)/2 , (2b+0)/2 ) = ( - a, b )
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4 years ago

Are all solutions roots?

White raven [17]

Hi there!

There shouldn't be roots in solutions. but sometimes if you use roots to simplify then roots can be in solutions.

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xXharleyquinn04Xx

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3 years ago

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Which of the following is a factor of 3x^3+18x^2+27x? A. 9x B. x^3 C. x+3 D. x-3

Tju [1.3M]

3x³ + 18x² + 27x = 3x(x² + 6x + 9) = 3x(x+3)²

Answer: C. x+3

3 0

3 years ago

Help thankss answer is <img src="https://tex.z-dn.net/?f=%20%28%20-%2014.0%29" id="TexFormula1" title=" ( - 14.0)" alt="

lyudmila [28]

We have point $P=(x_0,y_0)=(2,-4)$ , so first calculate $f'(x_0)$ . There will be:

$y=f(x)=2x^3-5x^2\\\\\\\\f'(x)=\big(2x^3-5x^2\big)'=\big(2x^3\big)'-\big(5x^2\big)'=2\big(x^3\big)'-5\big(x^2\big)'=\\\\=2\cdot3x^2-5\cdot2x=\boxed{6x^2-10x}\\\\\\\\f'(x_0)=f'(2)=6\cdot2^2-10\cdot2=6\cdot4-20=24-20=\boxed{4}$

Now, we can write the equation of the normal line as:

$y-y_0=-\dfrac{1}{f'(x_0)}(x-x_0)\\\\\\ y-(-4)=-\dfrac{1}{4}(x-2)\\\\\\y+4=-\dfrac{1}{4}x+\dfrac{1}{2}\\\\\\y=-\dfrac{1}{4}x+\dfrac{1}{2}-4\\\\\\\boxed{y=-\dfrac{1}{4}x-\dfrac{7}{2}}$

Normal line (and every line) crosses x-axis when y = 0, so coordinates of A:

$\boxed{y=0}\qquad\qquad\text{and}\\\\\\y=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\
0=-\dfrac{1}{4}x-\dfrac{7}{2}\\\\\\\dfrac{1}{4}x=-\dfrac{7}{2}\quad|\cdot4\\\\\\x=-\dfrac{7\cdot4}{2}\\\\\\x=-7\cdot2\\\\\\\boxed{x=-14}\\\\\\\boxed{A=(-14,0)}$

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3 years ago