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3 years ago

Which of the following exponential functions represents the graph below?

Mathematics

1 answer:

mario62 [17]3 years ago

3 0

Answer:

the graph corresponds to function "D" $f(x)=3\,(\frac{1}{5} )^x$

Step-by-step explanation:

Since the graph shown corresponds to an exponential "decay" (the function decreases as we move from left to right), the base of the exponent has to be a number smaller than 1 (one). So we examine the only two options that give such (options C and D which have fractions as the base - 1/3 and 1/5 respectively)

From there, we analyze which of the two functions satisfies the crossing of the y-axis at (0,3) which is clearly shown in the graph:

We study both:

function C at x = 0 gives:

$f(x)= 5\,(\frac{1}{3})^x\\f(0)=5\,(\frac{1}{3})^0 \\f(0)=5\,(1)\\f(0)=5$

while function D at x = 0 gives:

$f(x)= 3\,(\frac{1}{5})^x\\f(0)=3\,(\frac{1}{5})^0 \\f(0)=3\,(1)\\f(0)=3$

Therefore, the graph corresponds to function "D"

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1 year ago

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3 years ago

5 ÷ x for x = 1/3 solve

WARRIOR [948]

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2 years ago

How do u do b??????????

horrorfan [7]

the assumption being that the first machine is the one on the left-hand-side and the second is the one on the right-hand-side.

the input goes to the 1st machine and the output of that goes to the 2nd machine.

if she uses and input of 6 on the 2nd one, the result will be 6² - 6 = 30, if we feed that to the 1st one the result will be √( 30 - 5) = √25 = 5, so, simply having the machines swap places will work to get a final output of 5.

clearly we can never get an output of -5 from a square root, however we can from the quadratic one, the 2nd machine/equation.

let's check something, we need a -5 on the 2nd, so

$\bf \underset{final~out put}{\stackrel{y}{-5}}=x^2-6\implies 1=x^2\implies \sqrt{1}=x\implies 1=x$

so if we use a "1" as the output on the first machine, we should be able to find out what input we need, let's do that.

$\bf \underset{first~out put}{\stackrel{y}{1}}=\sqrt{x-5}\implies 1^2=x-5\implies 1=x-5\implies 6=x$

so if we use an input of 6 on the first machine, we should be able to get a -5 as final output from the 2nd machine.

$\bf \stackrel{first~machine}{y=\sqrt{\boxed{6}-5}}\implies y=\sqrt{1}\implies y=1 \\\\\\ \stackrel{second~machine}{y = \boxed{1}^2-6}\implies y = 1-6\implies y = -5$

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3 years ago