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alekssr [168]
3 years ago
6

Find two functions defined implicitly by the given relation y^2-64x^2=64

Mathematics
1 answer:
Reil [10]3 years ago
8 0

Answer: The two functions are

f(x) = sqrt(64x^2+64)

g(x) = -sqrt(64x^2+64)

Graphing these together forms a hyperbola

To get these two functions, we need to solve for y

y^2 - 64x^2 = 64

y^2 = 64x^2 + 64

y = sqrt(64x^2 + 64) or y = -sqrt(64x^2 + 64)

The last line is similar to how x^2 = 9 has two solutions (x = plus or minus 3)

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statuscvo [17]

Answer:

<em>Y = -4/3</em>

Step-by-step explanation:

Subtract 1 from -1/3.

Y = -1/3 - 1

Y = -1/3 - 3/3

Y = -4/3

6 0
3 years ago
Simplify the function. Than determine the key aspects of the function
Tom [10]

To simplify the function, we need to know some basic identities involving exponents.


1. b^(ax)=(b^x)^a=(b^a)^x

2. b^(x/d) = (b^x)^(1/d) = ((b^(1/d)^x)


Now simplify f(x), where

f(x)=(1/3)*(81)^(3*x/4)

=(1/3)(3^4)^(3*x/4) [ 81=3^4 ]

=(1/3)(3^(4*3*x/4) [ rule 1 above ]

=(1/3) (3^(3*x)

=(1/3)(3^(3x)) [ or (1/3)(27^x), by rule 1 ]



(A) Initial value is the value of the function when x=0, i.e.

initial value

= f(0)

=(1/3)(3^(3x))

=(1/3)(3^(3*0))

=(1/3)(3^0)

=(1/3)(1)

=1/3


(B) the simplified base base is 3 (or 27 if the other form is used)


(C) The domain for an exponential function is all real values ( - &infin; , + &infin; ).


(D) The range of an exponential function with a positive coefficient and without vertical shift is ( 0, + &infin; ).

8 0
3 years ago
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Nat2105 [25]
A number (we will call X) times 8 Is 8x
4 0
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PLEASE HELP!! EASY MATH FOR ADULTS NOT MIDDLE SCHOOLERS HELP PLS!
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Answer:

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C 2 and 3.2

Step-by-step explanation:

because anything lesser than the standard value of one is smaller.

because anything over the standard value of one is larger.

3 0
3 years ago
What is a possible situation for this graph
vichka [17]
Y=40x since y intercept is 0 and slope is 40
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Read 2 more answers
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