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3 years ago

6

Write the expression nine divided by the quantity of x plus y

1 answer:

seropon [69]3 years ago

7 0

9514 1404 393

Answer:

$\text{A)}\quad\dfrac{9}{x+y}$

Step-by-step explanation:

In the English expression, you need to understand that "A divided by B" means that A is the numerator and B is the denominator of the fraction representing the division. Here, "B" is "the quantity of ...", which means that the stuff after "of ..." is to be treated as a single item, the denominator in this case. That item is "x plus y." So, the expression in math terms is ...

$\dfrac{9}{x+y}$

When this is written in plain text, the "quantity" is put in parentheses, because the fraction bar no longer serves as a grouping symbol:

9/(x+y)

These expressions match the first choice, A.

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Find the perimeter of the image below.

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3 years ago

Write the equation of the line that contains the given point and has the given slope. (15, 31), slope is 0.5

Anna11 [10]

Y=.5x+23.5 is your answer :D

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4 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

A total of 327 tickets were sold for the school play. they were either adult tickets or student tickets. the number of student t

olga55 [171]

218 Students
109 Adults

Let x = adults
Let 2x = students

Add together and set equal to the total number.

3 0

3 years ago