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Schach [20]
3 years ago
5

Which of the following makes the statement true? 1/5^0 = ?a. 1 b. -1 c. -5 d. 0

Mathematics
2 answers:
irinina [24]3 years ago
5 0
Zero is your answer
natta225 [31]3 years ago
3 0
I hope this helps you



1/5^0


1/1


1
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What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .
VladimirAG [237]

Answer:

the  positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1

Using the standard form of the equation:

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term  indicating that the  hyperbola opens up and down.

a = distance that indicates  how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

y = \pm \frac{a}{b} (x-h)+k

where;

\frac{a}{b} is the slope

From above;

(h,k) = 11, 100

a^2 = 100

a = \sqrt{100}

a = 10

b^2 =4

b = \sqrt{4}

b = 2

y = \pm \frac{10}{2} (x-11)+2

y = \pm 5 (x-11)+2

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have  

\frac{a}{b} to be  the slope in the equation  y = \pm \frac{10}{2} (y-11)+2

\frac{a}{b}  = \frac{10}{2}

\frac{a}{b}  = 5

Thus, the  positive slope of the asymptote = 5

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Lilit [14]
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