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2 years ago

A rectangle has the vertices of 0,0 3,0 0,6 what is the area

Mathematics

1 answer:

laiz [17]2 years ago

5 0

The Answer i belive would be 18

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Abram rode in a cab for 9 miles. The taxi driver charged $2.50 for the first z

MAVERICK [17]

Answer:

Abram payed $12.50 for the cab ride in all.

7 0

2 years ago

30 pts AWNSER ASAP In 2 hours, Nolan can bake 5 1/2 dozen cookies. What is his unit rate in dozen cookies per hour?

seraphim [82]

Answer:

2 3/4 dozen cookies per hour

Step-by-step explanation:

To find his unit rate per hour, divide 5 1/2 by 2:

5.5/2

= 2.75 or 2 3/4 dozen cookies

So, his unit rate in dozen cookies per hour is 2 3/4

6 0

2 years ago

Problem Solving

Misha Larkins [42]

No. Four times five is twenty, so if Andrea has five five dollar bills, she can afford the tennis shoes. <span />

5 0

3 years ago

Given the function ƒ(x) = 20x2 − 18x − 25, find ƒ(−1).

dlinn [17]

f(x) = 20x² - 18x - 25

f(-1) = 20(-1)² - 18(-1) - 25

= 20 + 18 - 25

= 13

7 0

3 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