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3 years ago

What is the midpoint of the segment shown below?

Mathematics

2 answers:

Xelga [282]3 years ago

8 0

Answer:

4,3

Step-by-step explanation:

matrenka [14]3 years ago

5 0

Answer:

D. (4, 3)

Step-by-step explanation:

The total distance between the two points is 12 units. If you divide the total length by 2, which is 6, you can subtract 6 from the x value of point (10, 3) which is (4, 3) or you can add 6 to the x value of point (-2, 3) which is also (4, 3).

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An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d

denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. $w\times 2w\times h=36$

$h=\frac{18}{w^2}$

The equation form when top is open,

$f(w)=2w^2+2wh+2(2w)h$

Substitute the value of h,

$f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})$

$f(w)=2w^2+\frac{36}{w}+\frac{72}{w}$

$f(w)=2w^2+\frac{108}{w}$

Derivate w.r.t 'w',

$f'(w)=4w-\frac{108}{w^2}$

For critical point put it to zero,

$4w-\frac{108}{w^2}=0$

$4w=\frac{108}{w^2}$

$w^3=27$

$w^3=3^3$

$w=3$

Derivate the function again w.r.t 'w',

$f''(w)=4+\frac{216}{w^3}$

For w=3, $f''(3)=4+\frac{216}{3^3}=12 >0$

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = $\frac{18}{3^2}=2\ ft$

Therefore, the dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$

4 0

4 years ago

Dont be wrong got 2 more

valina [46]

Use the given formula
C = 2(pi)r
Given pi = 3.14, r = 4
C = 2(3.14)(4) = 25.12 inches

Solution: 25.12 inches

8 0

3 years ago

Read 2 more answers

Solve each expression using the correct order of operations,<br>7- 5 + 3 x 10<br>

Lynna [10]

Answer:

28 is your answer

Step-by-step explanation:

you start with multiplication and continue with addition and then go to subtraction

5 0

3 years ago

Read 2 more answers

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

Are the lines y = –x – 4 and 5x -5y = 20 perpendicular? Explain.

Zarrin [17]

Answer:

yes

Step-by-step explanation:

If 2 lines are perpendicular then the product of their slopes = - 1

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - x - 4 ← is in slope- intercept form

with slope m = - 1

5x - 5y = 20 ( subtract 5x from both sides )

- 5y = - 5x + 20 ( divide all terms by - 5 )

y = x - 4 ← in slope- intercept form

with slope m = 1

Thus product of their slopes is - 1 × 1 = - 1

Therefore the lines are perpendicular

7 0

3 years ago