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3 years ago

Y=1/6x+5 in standard form using integers.

2 answers:

6 0

In standard form the equation y=1/6x+5 would be
answer :x/6-y=-5
and if you have more problems like this here is a good website to go to it helps me alot
mathway is a good site to use

if you dont want to use its fine just trying to help :)

Brums [2.3K]3 years ago

6 0

In standard form, the equation would be:

x-6y=-30

This is because you cannot have fractions on standard form. So, I multiplied everything by 6 to get rid of the fraction. In standard form, you have all of the variables on one side of the equation. The formula for this is

Ax+By=C

This means x (and what is attached to x, A) always comes first. In this equation, A=1
B=-6
C=-30

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275 miles on 14 gallons<br><br><br>pls hurry

vaieri [72.5K]

Answer:

1 gallon per 19.64mi

Step-by-step explanation:

275/14 =19.6428571429

8 0

2 years ago

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How many rows and colums in 16 divided by 112

earnstyle [38]

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4 0

2 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

Help pleaseee............

stepan [7]

I say the answer is point c from the way the points are on the graph

3 0

3 years ago

Read 2 more answers

Graph the following piecewise function:

zhenek [66]

Check the picture below.

notice the solid circle at -4 and 3, since it includes those points.

also notice the dashed line, though the lines keeps on going those ways, is not included in the piece-wise, since it's not part fo the domain or constraints.

5 0

3 years ago