All categories

2 years ago

Which of the following exponential equations could be represented by the table below?

Mathematics

1 answer:

Nastasia [14]2 years ago

3 0

Answer:

Option 3: $y=(3)^{x-3}+2$

Step-by-step explanation:

See the attached figure.

We will make a table for each function of the options, then compare each one of them with the given table.

So, we can deduce that the function of option 3: $y=(3)^{x-3}+2$

is the exponential equations could be represented by the table.

Note both of the given table and $y=(3)^{x-3}+2$

has Horizontal Asymptote y =2

Which mean as x→-∞ ⇒ y→2

You might be interested in

Solve the inequality. Graph the solution.<br><br> w+6≤−3

yuradex [85]

Answer: w<-9 With the line underneath

Step-by-step explanation:

4 0

3 years ago

Last question please help

Grace [21]

Juan won and has a score of 8.5, the difference between there score is 3.5.

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

Does someone mind helping me with this problem? Thank you!

Damm [24]

Using the Pythagorean theorem:

c = sqrt(8^2 + 7^2)

c = sqrt(64 + 49)

c = sqrt(113)

c = 10.6

7 0

1 year ago

Read 2 more answers

The digit 4 in 49,308 has a value blank times as much as the 4 in 4,061

Step2247 [10]

Digit 4 in 49,308 is ten thousands and the 4in 4061 is thousands

8 0

3 years ago

Read 2 more answers