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2 years ago

Write the equation and slope intercept form y=mx+b

Mathematics

1 answer:

Nookie1986 [14]2 years ago

3 0

Here is your answer is slope-intercept form: y=-2x-15

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What is y=3x-14 in standard form

Dafna1 [17]

$The\ standard\ form:Ax+By=C\\\\=================\\\\y=3x-14\\\\3x-14=y\ \ \ \ \ \ |add\ 14\ to\ both\ sides\\\\3x=y+14\ \ \ \ \ \ |subtract\ y\ from\ both\ sides\\\\3x-y=14$

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2 years ago

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Pls help ill give branliest

Oksi-84 [34.3K]

Answer:

looks to me like it could be (A.)

Step-by-step explanation:

i might be wrong if i am im sorry

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2 years ago

The initial population of a town is 1,300 and it is increasing at the rate of 3% each year. Find the population at 12 years

stepan [7]

Answer:

1300 * (1 + 0.03)^12 = 1853

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2 years ago

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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

Log4 x + log4 (x – 3) = 1 Solve the equation.

Natasha_Volkova [10]

Answer:

x is EQUIVALENT to 2.89

Step-by-step explanation:

7 0

2 years ago