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Artyom0805 [142]

3 years ago

13

What is the distance between -6 and -3 on a number line?

2 answers:

Ira Lisetskai [31]3 years ago

7 0

I think it's 3 but idk I'm not good at math

d1i1m1o1n [39]3 years ago

5 0

I believe the distance betwee -6 and -3 on a number line is 3

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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Apply the rule (-x, y - 2) to the following point, what is the result?

BaLLatris [955]

Answer: Second Choice. (3, 3)

Step-by-step explanation:

<u>Given point</u>

(x, y) = (-3, 5)

<u>Given the rule for application</u>

(-x, y - 2)

<u>Determine the new x-value</u>

x = -3

-x = - (-3) = 3

<u>Determine the new y-value</u>

y = 5

y - 2 = (5) - 2 = 3

Therefore, the result is $\Large\boxed{(3,~3)}$

Hope this helps!! :)

Please let me know if you have any questions

5 0

2 years ago

Read 2 more answers

Which of the following is the statement of the triangle inequality theorem?

kupik [55]

we know that

<u>The triangle inequality theorem</u> states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side

so

Let

a,b,c------> the length sides of a triangle

The theorem states that three conditions must be met

<u>case 1)</u>

$a+b > c$

<u>case 2)</u>

$a+c > b$

<u>case3)</u>

$c+b > a$

therefore

<u>the answer is the option</u>

B. The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

5 0

4 years ago

Read 2 more answers

-3(x - 5) = 3x + 15

Travka [436]

Answer:

false

Step-by-step explanation:

because

-3(x-5)

-3x+15

6 0

3 years ago