Keesha will mow grass for $8 per hour write in standard form

<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

1 year ago

Select the correct answer. Twenty students in Class A and 20 students in Class B were asked how many hours they took to prepare

Lunna [17]

Answer:

B: The mean study time of students in Class B is less than students in Class A.

Step-by-step explanation:

To find out why answer B is the right answer, I will give you facts from each option.

Option A is false. The mean study time in Class A is 4.8. Meanwhile in Class B it is 4. For Class A, sum up the 20 study times which is 96 and divide them by 20, you will get 4.8 hours of mean study time. For Class B, the sum of the 20 study times is 80, which divided by 20 will be 4.

Option B is True. See previous explanation.

Option C is False. The median study time in Class B is 4. The median study time in Class A is 4.8,

Option D is False. The range in Class A is from 2 to 8. The range in Class B is from 2 to 7.

Option E is False: The mean and median study time of these classes is different.

6 0

2 years ago

Read 2 more answers

Need help with this math problem

zheka24 [161]

Answer:

m<1 = 51

m<2 = 61

m<3 = 29

Step-by-step explanation:

three angles of a triangle = 180

so knowing that add 68 + 61 to get 129

now subtract cause the last angle will = 180-129 = 51

for angle two is 61 cause their congruent

now we have one angle = 90 and another 61

add them 90+61= 151

now subtract 180-151 = 29

6 0

2 years ago

Given a = -3, b=4 and c= -5, evaluate a + b + c. 04 0-6 -12

Rainbow [258]

Answer: 0 - 4

Step-by-step explanation:

0 - 4 = -4

-3 + 4 = 1

1 - 5 = -4

6 0

2 years ago

Read 2 more answers

Y = 3 sin2x, y = 0, 0 ≤ x ≤ π; about the x−axis

Dominik [7]

I assume you're revolving the region with those bounds about the x-axis, and supposed to find the volume.

Via the disk method,

$\displaystyle\pi\int_0^\pi(3\sin2x)^2\,\mathrm dx=9\pi\int_0^\pi\sin^22x\,\mathrm dx$

Recall the half-angle identity for sine:

$\sin^2t=\dfrac{1-\cos2t}2$
$\implies\displaystyle\frac{9\pi}2\int_0^\pi(1-\cos4x)\,\mathrm dx$
$=\displaystyle\frac{9\pi}2\left(x-\frac14\sin4x\right)\bigg|_{x=0}^{x=\pi}$
$=\dfrac{9\pi^2}2$

8 0

2 years ago