What expression is equivalent to -3n-7+(-6n)+1

Kelsey can buy 2 pounds of apples for 7$. Write an equation to represent the cost, c, for a, pounds of apples.

zmey [24]

$\huge\boxed{c=3.5a}$

First, we need to find how much a single pound of apples costs.

Do this by dividing.

$\frac{\$7}{2\ \text{lbs.}}\div\frac{2}{2}=\frac{\$(7\div 2)}{(2\div 2)\ \text{lbs.}} = \frac{\boxed{\$3.50}}{1\ \text{lb.}}$

Now, write an equation.

$\large\boxed{c=3.5a}$

This represents that the cost equals 3.5 times the number of apples.

8 0

3 years ago

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In 2000, there were 12900 students at college A, with a projected enrollment increase of 900 students per year. In the same year

LenaWriter [7]

Answer:

Step-by-step explanation:

Let x represent the number of years it will take the two colleges to have the same enrollment.

In 2000, there were 12900 students at college A, with a projected enrollment increase of 900 students per year. This means that the expected number of students at college A in x years time is

12900 + 900x

In the same year, there were 25,000 students at college B, with a projected enrollment decline of 700 students per year. This means that the expected number of students at college B in x years time is

25000 - 700x

For both colleges to have the same enrollment,

12900 + 900x = 25000 - 700x

900x + 700x = 25000 - 12900

1600x = 12100

x = 12100/1600

x = 7.56

Approximately 8 years

The year would be 2000 + 8 = 2008

4 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

How many solutions does 8x = 56 have?

OlgaM077 [116]

There is only one answer, x=7

3 0

3 years ago

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I NEED HELP PLEASE (BE SERIOUS AND RIGHT ANSWERS ONLy)

True [87]

567 x 0.20 = 113.4
Closest Answer = 110

5 0

3 years ago

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