All categories

2 years ago

Aquarium I contains 36.4 gallons of water. Aurora will begin filling Aquarium I at a rate of 1.9 gallons per minute.

Mathematics

2 answers:

dybincka [34]2 years ago

5 0

Answer:

answer is d

Step-by-step explanation:

i did the quiz

Vladimir [108]2 years ago

5 0

Ummmm The answer is D

You might be interested in

What is the correct formula for density

zvonat [6]

Answer:

C. density = mass / volume

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the value of x?

STatiana [176]

50

Step by step explanation

4 0

2 years ago

Read 2 more answers

Please help ASAP!!!!

Genrish500 [490]

Answer:

is B

Step-by-step explanation:

4 0

2 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 23 in. by 13 in. by

faltersainse [42]

Answer:

The dimension of the box is 17.66 in by 7.66 in by 2.67 in.

Therefore the volume of the box is 361.19 $in^3$ .

Step-by-step explanation:

Given that the dimensions of a cardboard is 23 in by 13 in.

Let the side of the square be x in.

Then the length of the box= (23-2x) in

and the width of the box =(13-2x) in

and height = x in.

The volume of the box is = length ×width × height

=[(23-2x)(13-2x)x] $in^3$

=(299x-72x² +4 $x^3$ ) $in^3$

∴V=299x-72x² +4x³

Differentiating with respect x

V'= 299-144x+12x²

Again differentiating with respect x

V''= -144+24x

To find the dimensions, we set V'=0

∴299-144x+12x²=0

Applying Sridharacharya formula that is the solution of a quadratic equation ax²+bx+c is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here a=12, b=-144 , c=299

$\therefore x=\frac{-(-144)\pm\sqrt{(-144)^2-4.12.299}}{2.12}$

$\Rightarrow x= 9.33, 2.67$

If we take x=9.33 in, then the width of the box [13-(2×9.33)] will negative.

∴x = 2.67 in

If at x = 2.67, V''<0 , then the volume of the box will be maximum or V''>0 then volume of the box will be minimum.

$V''|_{x=2.67}=-144+(24\times 2.67)=-79.92$

Therefore at x = 2.67, the volume of the box maximum.

The length of the box =[23-(2×2.67)] in

=17.66 in

The width of the box =[13-(2×2.67)] in

=7.66 in

The height of the box= 2.67 in

The dimension of the box is 17.66 in by 7.66 in by 2.67 in.

Therefore the volume of the box is =(17.66×7.66×2.67) $in^3$

=361.19 $in^3$

8 0

2 years ago

Ples help me find slant assemtotes

FrozenT [24]

A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

$(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}$

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

$2x-1+2x-ax-b=(4-a)x-(b+1)$

which tells us that we would choose $a=4$ . You might be tempted to think $b=-1$ , but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of $b$ , we have to solve for it in the following limit.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

$-\dfrac12=\dfrac{b+1}2\implies b=-2$

So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

Reduce the "negligible" term to get

$\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0$

Now we take $a=0$ , and again we're careful to not pick $b=-1$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0$

$\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2$

$(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}$

This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

4 0

3 years ago