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HACTEHA [7]
4 years ago
7

The graph of an exponential function is given. Which of the following is the correct equation of the function? (2 points)

Mathematics
2 answers:
adell [148]4 years ago
6 0

Answer:

Option C

Step-by-step explanation:

WE are given a graph of exponential function and asked to find out of four functions which function the graph represents.

We find that any function of the form

y=a^x

passes through (0,1) for all a

Hence we cannot distinguish any thing from the point

Let us consider when x=1

Then first function will have point as (1,2.4)

II funciton (1,0.31)

III function (1,0.45)

IV funciton (1,1.8)

From the graph given we find that the point of y is less than 1.  Hence it can be either II or III

Consider x=-1

Then II function has point as (-1,3.225) and III function has point as (-1,2.222)

From the graph we find that the graph satsified only the point (-1,2.2222)

Hence OPtion III is right

IgorLugansk [536]4 years ago
3 0

Answer:

y=0.31^{x}

B is correct

Step-by-step explanation:

We are given exponential function graph. We need to choose correct equation of exponential function.

in graph we can see start from II quadrant and end in I quadrant.

It would be decay exponential function.

y=ab^x

b must be less than for given graph.

y-intercept of given graph is (0,1)

a=1, because y-intercept gives initial value of graph.

0<b<1

In the given option B and C whose b is less than 1.

Now, we take another point x=-1

For x=-1, the value of y of given graph between 3 and 4

Now we find value of y for x=-1

For option B: y=0.31^{-1}=3.2

For option C: y=0.45^{-1}=2.2

Hence, B is correct equation for given graph.

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What is the value of log0.5 16 (0.5 is the small bottom number)<br>-4.00<br>-0.25<br>1.51<br>2.41
eduard

we are given

log_0_._5(16)

Firstly, we will factor out 16

16=2\times 2\times 2\times 2\times 2

16=2^4

we can write it as

16=(\frac{1}{2})^{-4}=(0.5)^{-4}

we can replace it as

log_0_._5(16)=log_0_._5((\frac{1}{2})^{-4})

now, we can use property of log

log_a(b^n)=nlog_a(b)

we get

log_0_._5(16)=-4log_0_._5(0.5)

log_0_._5(16)=-4\times 1

log_0_._5(16)=-4.............Answer


4 0
3 years ago
Read 2 more answers
. Equilibrium: If the number of persons who attend the club meeting this week is X, then the number of people who will attend ne
Artyom0805 [142]

Answer:

100

Step-by-step explanation:

In economics, for a firm to earn optimum profits, it is important that it achieves a long run equilibrium. We can transfer the same to the case here that for the club to achieve optimum attendance, it must achieve long- run equilibrium attendance.

The condition for Long Run Equilibrium is that:

Club meeting attendance this week = Club meeting attendance next week

X = 80 + 0.20X

X - 0.20X = 80

X = 80/0.8

X = 100.

The long- run equilibrium attendance for this club is 100.

5 0
4 years ago
1A research firm finds that the average number of customers that are in Somerset Mall on a
SpyIntel [72]

Answer: C(t) = -475.5*cos(t*pi/6) + 475.5

Step-by-step explanation:

We know that we have a sinusoidal relation, with a minimum at 9:00 am and at 9:00 pm.

If we define the 9:00 am as our t = 0, we have that the maximum, at 3:00pm, is at t = 6 hours.

and the other minimum, at 9:00pm, is at t = 12 hours.

Then we need to find a trigonometric function that has the minimum at t = 0, we can do this as:

-Cos(c*t)

when t = 0

-cos(0) = - 1

then we have a function:

C(t) = -A*cos(c*t) + B

where A, c and B are constants.

We know that at t = 0 we have 0 customers, and that at t = 6h we have 875 customers, that is the maximum.

then:

C(0) = 0 = -A + B

this means that A  = B, then our function is:

C(0) = -A*cos(c*t) + A.

now, at t = 6h we have a maximum, this means that -A*cos(c*6h) = A

then:

C(6h) = A + A = 875

2A = 875

A = 875/2 = 437.5

and we also have that, if -cos(c*6) = 1

then cos(c*6) = -1

and we know that cos(pi) = -1

then c*6 = pi

c = pi/6

Then our function is

C(t) = -475.5*cos(t*pi/6) + 475.5

8 0
3 years ago
Graphs can be used to estimate information not represented on the graph. True or False
Verizon [17]
I believe that it is true.
5 0
3 years ago
Read 2 more answers
Q1 A ball is thrown upwards with some initial speed. It goes up to a height of 19.6m and then returns. Find (a) The initial spee
lubasha [3.4K]

Answer:

(a)  19.6 ms⁻¹

(b)  2 s

(c)  9.8 ms⁻¹

(d)  4 s

Step-by-step explanation:

<u>Constant Acceleration Equations (SUVAT)</u>

\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}

When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.

Acceleration due to gravity = 9.8 ms⁻².

<h3><u>Part (a)</u></h3>

When the ball reaches its maximum height, its velocity will momentarily be zero.

<u>Given values</u> (taking up as positive):

s=19.6 \quad v=0 \quad a=-9.8

\begin{aligned}\textsf{Using} \quad v^2&=u^2+2as\\\\\textsf{Substitute the given values:}\\0^2&=u^2+2(-9.8)(19.6)\\0&=u^2-384.16\\u^2&=384.16\\u&=\sqrt{384.16}\\\implies u&=19.6\; \sf ms^{-1}\end{aligned}

Therefore, the initial speed is 19.6 ms⁻¹.

<h3><u>Part (b)</u></h3>

Using the same values as for part (a):

\begin{aligned}\textsf{Using} \quad s&=vt-\dfrac{1}{2}at^2\\\\\textsf{Substitute the given values:}\\19.6&=0(t)-\dfrac{1}{2}(-9.8)t^2\\19.6&=4.9t^2\\t^2&=\dfrac{19.6}{4.9}\\t^2&=4\\t&=\sqrt{4}\\\implies t&=2\; \sf s\end{aligned}

Therefore, the time taken to reach the highest point is 2 seconds.

<h3><u>Part (c)</u></h3>

As the ball reaches its maximum height at 2 seconds, one second before this time is 1 s.

<u>Given values</u> (taking up as positive):

u=19.6 \quad a=-9.8 \quad t=1

\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=19.6+(-9.8)(1)\\v&=19.6-9.8\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}

The velocity of the ball one second before it reaches its maximum height is the <u>same</u> as the velocity one second after.

<u>Proof</u>

When the ball reaches its maximum height, its velocity is zero.

Therefore, the values for the downwards journey (from when it reaches its maximum height):

u=0 \quad a=9.8 \quad t=1

(acceleration is now positive as we are taking ↓ as positive).

\begin{aligned}\textsf{Using} \quad v&=u+at\\\\\textsf{Substitute the given values:}\\v&=0+9.8(1)\\\implies v&=9.8\; \sf ms^{-1}\end{aligned}

Therefore, the velocity of the ball one second before <u>and</u> one second after it reaches the maximum height is 9.8 ms⁻¹.

<h3><u>Part (d)</u></h3>

From part (a) we know that the time taken to reach the highest point is 2 seconds.  Therefore, the time taken by the ball to travel from the highest point to its original position will also be 2 seconds.

Therefore, the total time taken by the ball to return to its original position after it is thrown upwards is 4 seconds.

4 0
1 year ago
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