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Kryger [21]
3 years ago
5

Simplify each expression t+8t

Mathematics
2 answers:
Schach [20]3 years ago
7 0
Answer is 9t......................................
FromTheMoon [43]3 years ago
5 0
Think of it this way. Let's say t= one. 1+8 is 9. Your answer is 9t. Think of a variable without a number as one. 
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Match the following
Roman55 [17]

Answer:

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Step-by-step explanation:

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2 years ago
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Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
zavuch27 [327]
Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }
\\\\
\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h,  k\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2 + b ^2}\\
asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)
\end{cases}\\\\
-------------------------------

\bf \begin{cases}
h=2\\
k=-3\\
a=6\\
c=2\sqrt{10}
\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1
\\\\\\
c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2
\\\\\\
4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2
\\\\\\
\sqrt{4}=b\implies 2=b\\\\
-------------------------------\\\\
\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
3 years ago
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fredd [130]

Answer:

Multiply 2 by the coefficient 4 and you get 8x, which is the answer

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Fed [463]

algebra.............

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Determine the zero solution<br> f(x)=3x^(2)+5x-12
RideAnS [48]

Answer:

x= -3 or 4/3

Step-by-step explanation:

3x²+5x-12 =0

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3x(x+3) -4(x+3) =0

(3x-4)(x+3)=0

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