All categories

2 years ago

This should contain only positive exponents

Mathematics

1 answer:

Simora [160]2 years ago

4 0

(4a³)
-------
(8a³)

Since the variable and exponent are exactly the same, and you're dividing them, they cancel out.

(4)
----
(8)

This can be simplified to 1/2.

You might be interested in

Solve the following system of equations below x-2y=3 2x-3y=9

Artyom0805 [142]

$x - 2y = 3 \\ 2x - 3y = 9 \\ \\ - 2x + 4y = - 12 \\ \: 2x - 3y = 9 \\ \\ y = - 3 \\ 2x + 9 = 9 \\ 2x = 0 \\ x = 0 \\ x = 0 \: \: y = - 3$

7 0

3 years ago

Read 2 more answers

What is the slope of a line that is perpendicular to a line whose equation is 3y=−4x+2 ?

charle [14.2K]

The slope of the given line is -4/3. the slope of any perpendicular line is the negative reciprocals, or 3/4

8 0

3 years ago

Please help <br><br> Maths....

lora16 [44]

6 cm from what im seeing

6 0

3 years ago

Read 2 more answers

Please find the slope

Morgarella [4.7K]

The answer to your question is A: 5/3. To get to the top point, you have to go up 5, then go 3 to the right.

4 0

3 years ago

Read 2 more answers

<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