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3 years ago

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Martell had boxes he was stacking.Each was 2 feet high.If he stacks 3 boxes on top of a table that is 3 feet high,what will be t

he total height of the boxes and the table?

2 answers:

padilas [110]3 years ago

7 0

5 because the table is 3 feet itself plus 2 more feet is 5

Lunna [17]3 years ago

4 0

9 feet

(2*3)+3
(6)+3
=9

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Circle P has a circumference of approximately 75 inches. Circle P with radius r is shown. What is the approximate length of the

Komok [63]

Answer:

12 inches, I hope this is right! :)

Step-by-step explanation:

circumference = 75 inches

75 / 2 = 35.7

37.5 / π = 11.9366207

Rounded to the nearest inch:

12 inches

8 0

3 years ago

Read 2 more answers

The product of consecutive odd integers is 575. what is the largest integer? 25 23 525 need more information

hichkok12 [17]

sqrt(575) = 23.979

so 23 is one number

575/23 = 25

the two numbers would be 23 * 25

25 is the largest one

4 0

3 years ago

Evaluate the function f(x)=x^2+2x+3 at the given values of the independent variable and simplivy

tatyana61 [14]

Answer: i) f(-9)= 66

ii) $f(x+1)=x^2+4x+6$ .

iii) $f(-x)=x^2-2x+3$

Step-by-step explanation:

The given function : $f(x)=x^2+2x+3$ .

i) For f(-9) , the independent variable is x= -9.

Put x= -9 in given function , we get

$f(-9)=(-9)^2+2(-9)+3$

$=81-18+3=66$

Thus , f(-9)= 66

ii) For f(x+1) , the independent variable is x= x+1.

Replace x by x+1 in given function , we get

$f(x+1)=(x+1)^2+2(x+1)+3$ .

$=(x^2+1+2x)+2x+2+3$ $[\because\ (a+b)^2=(a^2+b^2+2ab)]$

$=x^2+1+2x+2x+5$

$=x^2+4x+6$

Thus , $f(x+1)=x^2+4x+6$ .

iii)For f(-x) , the independent variable is x=-x.

Replace x by -x in given function , we get

$f(-x)=(-x)^2+2(-x)+3$ .

$=x^2-2x+3$ [∵ (+)(-)=(-)]

Thus , $f(-x)=x^2-2x+3$

8 0

2 years ago

What is the slope of a line that goes through the points (2, 5) and (8, 6) ?

jenyasd209 [6]

Answer:

m = 1 / 6

Step-by-step explanation:

Use this formula y2 - y1 / x2 -x1 to find the slope.

6 - 5 / 8 -2 = 1/6

8 0

2 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