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Rainbow [258]
3 years ago
12

The circles centered at points A, C, and D have radii of length AC.

Mathematics
1 answer:
natita [175]3 years ago
3 0

it is the second one

Step-by-step explanation:

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What is the cirumference of a circle with a diameter of 4.1 cm
nordsb [41]

<em>Answer:</em>

<em>C = 12.88</em>

<em>Step-by-step explanation:</em>

<em>If the radius of the circle is 4 cm, then the diameter of the circle is 4 cm x 2, or 8 cm. If you know the circumference of the circle, divide it by </em><em>π</em><em> to get the diameter.</em>

<em>C = π d = π · 4.1 = </em><em>12.88</em><em>053cm </em>

Glad to help ya!!

5 0
3 years ago
6. Write an equation of a line parallel to y = 3x + 9 and goes through the point (-1,5).
lisabon 2012 [21]

Answer: The equation parallel to the line is y=3x+8

Step-by-step explanation: Parallel lines have the same gradients, wr have a gradient for the above line which 3. The standard formula for the linear equation is y=mx+c

We have to calculate c since we have x and y.

y=mx+c

5= 3(-1)+c

c=8

The equation become y=3x+8

4 0
3 years ago
I need help ASAP!
sertanlavr [38]
Sin (x) = 12/13 this is because sine is opposite divided by hypotenuse.
cos (x) = 5/13 this is because cosine is adjacent divided by hypotenuse.
tan (x) = 12/5 this is because tangent is opposite divided by adjacent.
3 0
3 years ago
Simplify the expression.<br> 0.24 – 1.6
lukranit [14]

Answer:

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Determine the equations of the vertical and horizontal asymptotes, if any, for y=x^3/(x-2)^4
djverab [1.8K]

Answer:

Option a)

Step-by-step explanation:

To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

\lim_{x\to\\2}\frac{x^3}{(x-2)^4} \\\\\\lim_{x\to\\2}\frac{2^3}{(2-2)^4}\\\\\lim_{x\to\\2}\frac{2^3}{(0)^4} = \infty

Then. x = 2 it's a vertical asintota.

To obtain the horizontal asymptote of the function take the following limit:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4}

if \lim_{x \to \infty}\frac{x^3}{(x-2)^4} = b then y = b is horizontal asymptote

Then:

\lim_{x \to \infty}\frac{x^3}{(x-2)^4} \\\\\\lim_{x \to \infty}\frac{1}{(\infty)} = 0

Therefore y = 0 is a horizontal asymptote of f(x).

Then the correct answer is the option a) x = 2, y = 0

3 0
3 years ago
Read 2 more answers
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