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KiRa [710]
2 years ago
6

Simplify and answer the boxes.

Mathematics
1 answer:
s2008m [1.1K]2 years ago
6 0

Answer:

\huge\boxed{\dfrac{x^2+9^2}{x-3y}+\dfrac{6xy}{3y-x}=x-3y}

Step-by-step explanation:

Domain:

x-3y\neq0\Rightarrow x\neq3y

\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{3y-x}=\dfrac{x^2+9y^2}{x-3y}+\dfrac{6xy}{-(x-3y)}\\\\=\dfrac{x^2+9y^2}{x-3y}-\dfrac{6xy}{x-3y}=\dfrac{x^2+9y^2-6xy}{x-3y}\\\\=\dfrac{x^2-2(x)(3y)+(3y)^2}{3y-x}=\dfrac{(x-3y)^2}{3y-x}\\\\=\dfrac{\bigg[-1(3y-x)\bigg]^2}{3y-x}=\dfrac{(-1)^2(3y-x)^2}{3y-x}\\\\=\dfrac{1(x-3y)(x-3y)}{x-3y}=x-3y

Used:

The distributive property: a(b + c) = ab + ac

(a - b)² = a² - 2ab + b²

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Which of the segments below is a secant?
Rainbow [258]

Answer:

BC

Step-by-step explanation:

A secant is a line which intersects the circle at 2 points

In the given diagram this is BC

3 0
3 years ago
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Plzzz helpppppp!!!!!!
Tema [17]

Answer:

(c) h(3) = 1

Step-by-step explanation:

Given

The attached graph

Required

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The formula to use in solving this is:

y = h(x)

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y = h(3)

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2 years ago
5)
Paladinen [302]

Answer:183.4

Step-by-step explanation:

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6 0
3 years ago
I don't know if this is right... please someone help mee
worty [1.4K]
For the first circle, let's use the pythagorean theorem

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{8^2+15^2}\implies c=\sqrt{289}\implies c=17

now, it just so happen that the hypotenuse on that triangle, is actually 17, but we used the pythagorean theorem to find it, and the pythagorean theorem only works for right-triangles.

 so if the hypotenuse is actually 17, that means that triangle there is actually a right-triangle, meaning that the radius there, and the outside line there, are both meeting at a right-angle.

when an outside line touches the radius line, and they form a right-angle, the outside line is indeed a tangent line, since the point of tangency is always a right-angle with the radius.



now, let's check for second circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{11^2+14^2}\implies c=\sqrt{317}\implies c\approx 17.8044938

well, low and behold, we didn't get our hypotenuse as 16 after all, meaning, that triangle is NOT a right-triangle, and that outside line is not touching the radius at a right-angle, therefore is NOT a tangent line.



let's check the third circle

\bf \textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies c=\sqrt{a^2+b^2}
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
c=\sqrt{33^2+56^2}\implies c=\sqrt{4225}\implies c=\stackrel{33+32}{65}

this time, we did get our hypotenuse to 65, the triangle is a right-triangle, so the outside line is indeed a tangent line.
6 0
2 years ago
Ron has $49.Matt has $19 more than Ron.Naomi has $12 less than Matt.How much money do they have altogether?
EleoNora [17]

The answer to your question is,

They have a total of $173 altogether!

-Mabel <3

5 0
2 years ago
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