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Tema [17]
3 years ago
6

What is a equivalent fraction for 3/9?

Mathematics
1 answer:
kherson [118]3 years ago
5 0
1/3 = 2/6 = 3/9 = 4/12 = 5/15 = 6/18 = 7/21 = 8/24 = 9/27 = 10/30 = 11/33 = 12/36 = 13/39 = 14/42 = 15/45 = 16/48 = 17/51 = 18/54 = 19/57 = 20/60
You can use any, these are just the smallest to the biggest but any are fine but I would use the first one. Hope I helped. Equivalent is what 3/9 equals to.
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find the equation of the circle where (-9,4),(-2,5),(-8,-3),(-1,-2) are the vertices of an inscribed square.
solniwko [45]
Check the picture below, so, that'd be the square inscribed in the circle.

so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad 
%  (c,d)
&({{ -8}}\quad ,&{{ -3}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
\stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2}
\\\\\\
d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2}
\\\\\\
d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10

that means the radius r = 5.

now, what's the center?  well, the Midpoint of the diagonals, is really the center of the circle, let's check,

\bf \textit{middle point of 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ -2}}\quad ,&{{ 5}})\quad 
%  (c,d)
&({{ -8}}\quad ,&{{ -3}})
\end{array}\qquad 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{-8-2}{2}~,~\cfrac{-3+5}{2} \right)\implies (-5~,~1)

so, now we know the center coordinates and the radius, let's plug them in,

\bf \textit{equation of a circle}\\\\ 
(x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2
\qquad 
\begin{array}{lllll}
center\ (&{{ h}},&{{ k}})\qquad 
radius=&{{ r}}\\
&-5&1&5
\end{array}
\\\\\\\
[x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25

8 0
3 years ago
Using the figure below, select the pairs of complementary angles.
Mila [183]

Answer:

c) <DBE and <EBC

Step-by-step explanation:

This is because when added together, both angles equal 90 degrees.

Complementary angles always equal 90 degrees.

8 0
3 years ago
Please help I need this please help please just me help
Nikitich [7]

Answer:

It would take 4 minutes.

Step-by-step explanation:

Step 1: divide 63 by 3 to get how many hotdogs the champion can eat in a minute which equals 21

Step 2: Divide 84 by 21 to get 4 minutes

3 0
2 years ago
The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Th
Fantom [35]

Answer:

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C

Step-by-step explanation:

To find:

∫6x^5(x^6-2)\,dx

Solution:

Method of substitution:

Let x^6-2=t

Differentiate both sides with respect to t

6x^5\,dx=dt

[use (x^n)'=nx^{n-1}]

So,

∫6x^5(x^6-2)\,dx = ∫ t\,dt = \frac{t^2}{2}+C_1 where C_1 is a variable.

(Use ∫t^n\,dt=\frac{t^{n+1} }{n+1} )

Put t=x^6-2

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1

Use (a-b)^2=a^2+b^2-2ab

So,

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C

where C=2+C_1

Without using substitution:

∫6x^5(x^6-2)\,dx = ∫6x^{11}-12x^5\,dx = \frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C

So, same answer is obtained in both the cases.

7 0
2 years ago
If u help I’ll give u brainliest but please give correct answer
Black_prince [1.1K]

Answer:

here is no question sorry

4 0
3 years ago
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