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3 years ago

Complete the steps to add 3 2/5 + 1 3/10

Mathematics

2 answers:

Nimfa-mama [501]3 years ago

6 0

Answer:

1.C is correct answer and 2.C and 3.B is correct answer.

frozen [14]3 years ago

4 0

The answers are C C and B

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A company sells its printers to customers in order to make a profit of 25% calculate the price the company paid for a printer wh

ioda

The company paid $2000 for the printer

Step-by-step explanation:

Let x be the price of the printer for which the company bought it.

If x is the price of printer, then 25% of x will be the profit made by the company.

That is:

0.25x

According to the statement,

The cost and profit earned by company are equal to 2500

So,

$x+0.25x = 2500\\1.25x = 2500$

Dividing both sides by 1.25

$\frac{1.25x}{1.25} = \frac{2500}{1.25}\\x = 2000$

Hence,

The company paid $2000 for the printer

Keywords: Percentage, Profit

Learn more about percentage at:

#LearnwithBrainly

7 0

3 years ago

Which of the following statements are true about the lengths of the trails? Check all that apply.

SVEN [57.7K]

Where are the options?

3 0

3 years ago

The drama club you are part of is hosting a pancake breakfast. You are in charge of buying the sausage, which costs $5 per pound

antiseptic1488 [7]

Answer:

Step-by-step explanation:

Look at the pictures.

x - sausage

y - bacon

A. 20 pounds of sausage and 90 pound of bacon

x = 20 → y = 100 > 90

B. 40 pound of sausage and 40 pound of bacon

x = 40 → y = 75 > 40

C. 60 pound of sausage and 80 pound of bacon

x = 60 → y = 50 < 80

D. 80 pound of sausage and 20 pound of bacon

x = 80 → y = 25 > 20

4 0

3 years ago

Read 2 more answers

Determine the center and radius of the following circle equation:

-BARSIC- [3]

Answer:

(6, 9 ) and r = 3

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Given

x² + y² - 12x - 18y + 108 = 0

Rearrange the x- terms and the y- terms together and subtract 108 from both sides, that is

x² - 12x + y² - 18y = - 108

To obtain standard form use the method of completing the square

add ( half the coefficient of the x and y terms )² to both sides

x² + 2(- 6)x + 36 + y² + 2(- 9)y + 81 = - 108 + 36 + 81

(x - 6)² + (y - 9)² = 9 ← in standard form

with centre = (6, 9 ) and r = $\sqrt{9}$ = 3

7 0

3 years ago

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