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ivann1987 [24]
2 years ago
15

Which is the graph of f(x) = one-fourth(4)x?

Mathematics
2 answers:
frez [133]2 years ago
6 0

Answer:

The image shows the graph for given function.    

Step-by-step explanation:

We are given the following information in the question:

f(x) = \displaystyle\frac{1}{4}4^x

It is clear function is an exponential function and have shape similar to exponential function.

An exponential function is of the form:

f(x) = b^x,

where b is a parameter of the function and read as b raised to the power x.

The exponential function enjoys the following properties:

  • If  0 < b < 1, then the graph decreases as we move from left to right.
  • If b > 1, then the graph will increase as we move from left to right.

matrenka [14]2 years ago
4 0

Answer:

it is the third graph on edge

Step-by-step explanation:

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The equation of a circle is (x + 12)2 + (y + 16)2 = (r1)2, and the circle passes through the origin. The equation of the circle
Furkat [3]
For an equation to pass through the origin, (0,0) must be a solution of the equation. So, for the first one, 

(x + 12)^{2} + (y + 16)^{2} = r_{1}^{2}

(0, 0) must satisfy the equation and we can solve for the value of r₁ as shown below.

(0 + 12)^{2} + (0 + 16)^{2} = r_{1}^{2}
12^{2} + 16^{2} = r_{1}^{2}
r_{1} = \sqrt{12^{2} + 16^{2}}
r_{1} = 20

Since the second equation of the circle also passes through the origin, we use the same steps from the first one.

(0 - 30)^{2} + (0 - 16)^{2} = r_{2}^{2}
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r_{2} = 34

Thus, we have the values for r₁ and r₂ as 20 and 34, respectively.

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2 years ago
17. A local university accepted 2,300 students out of 4,500 applicants for admission. What was the acceptance rate, expressed as
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2 years ago
The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t
Doss [256]

Answer:

First Case:

\displaystyle p=\frac{5}{2}\text{ and } d=-2

Second Case:

\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

6p+2, 4p^2-10, \text{ and } 4p+3

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

4p^2-10=(6p+2)+d

And, likewise, for the third term:

4p+3=(6p+2)+2d

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

d=4p^2-6p-12

And for the second equation:

2d=-2p+1

To avoid fractions, let's multiply the first equation by 2. Hence:

2d=8p^2-12p-24

Therefore:

8p^2-12p-24=-2p+1

Simplifying yields:

8p^2-10p-25=0

Solve for <em>p</em>. We can factor:

8p^2+10p-20p-25=0

Factor:

2p(4p+5)-5(4p+5)=0

Grouping:

(2p-5)(4p+5)=0

Zero Product Property:

\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}

Then, we can use the second equation to solve for <em>d</em>. So:

2d_1=-2p_1+1

Substituting:

\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

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Answer:

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