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2 years ago

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Small paper clips cost $1.20 per package of 150. The ratio of the number of large paper clips needed to fill the allotted space

is 9.5. Compare the total cost of the small paper clips to the large paper clips

1 answer:

Nikitich [7]2 years ago

3 0

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Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

Deedra has the equations for lines A and B. When she solved for the point where these two lines intersect, she ended up with the

KonstantinChe [14]

we can take a peek at two of those lines hmmm say y = 5x + 3 and y = 5x + 7.

let's notice, those two equations for those lines are in slope-intercept form, so let's solve the system.

since y = y then

5x + 3 = 5x + 7

3 = 7 what the?

well, notice, both lines have the same slope of 5, but different y-intercept, one has it at y = 3 and the other at y = 7, what does that mean?

it means that both lines are parallel to each other, one may well be above the other, but both are parallel, and since a solution to the system is where their graphs intersect, well, parallel lines never touch, so a system with two parallel lines has no solutions.

3 0

3 years ago

Read 2 more answers

The ratio of boys to girls in Mrs. Miller's class is 2:3. If there are 8 boys in her class, how many girls are there?

Alenkasestr [34]

Answer:12

for every 2 boys that is 3 girls

there are 4 sets of two boys so you would have 4 sets of 3 girls

4×3=12

I hope this is good enough:

7 0

2 years ago

I have 3 questions for math

Airida [17]

Answer:

no question content
(x, y) = (1/2, 4)
(x, y) = (2, 10)

Step-by-step explanation:

1. No link, no question

__

2. Divide the first equation by 2 and substitute for y using the expression in the second equation.

2x +(4x+2) = 5

6x = 3 . . . . . . . . . subtract 2

x = 3/6 = 1/2 . . . . divide by2

y = 4(1/2) +2 = 4 . . . . substitute into the equation for y

The solution is (x, y) = (1/2, 4).

__

3. The solution using a graphing calculator is (x, y) = (2, 10). (see attached)

7 0

2 years ago

Is 23.044 greater than -23.060

My name is Ann [436]

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Yes, it is.

Positive numbers are greater than negative numbers.

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

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3 0

2 years ago

Read 2 more answers