Which equation describes the same line as y

Simplify (4n^4 - 3n^2 + n)+ (2n^4 -7n^2 + 9n). Express your answer without factoring.

Pie

Answer:

6n^4-10n^2+10n

Step-by-step explanation:

4 n^4+2n^4-3n^2-7n^2+9n+n

6n^4-10n^2+10n

4 0

3 years ago

Read 2 more answers

Find last year’s salary if, after a 3% pay raise, this year’s salary is $30,900

Liono4ka [1.6K]

Your answer is $30,000.
The way I have answered this is quite strange, but I'll do my best to explain it. So because we know that $30,900 is 3% than last year, we can call it 103%. This allows us to form a ratio and therefore find 100%.
30,900 : 103
÷ 103
300 : 1
× 100
30,000 : 100
Which means $30,000 is 100%, or 3% less than $30,900. I hope this helps! Let me know if it was confusing or anything :)

5 0

3 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

PLEASE HELP ASAP Due in 1 day!!!

Helen [10]

Step-by-step explanation:

Q1 . (f+g)(x) = f(x) + g(x)

=4x-4 +2x^2 -3x

= 2x^2 + x -4

Q2. (f-g)(x) = f(x) - g(x)

= 2x^2−2 - (4x+1)

= 2x^2 -2 -4x -1

= 2x^2 - 4x -3

Q3. h(x)=3x−3 and g(x)=x^2+3

(h.g)(x) = h(x) × g(x)

= (3x-3) × (x^2 + 3)

=3x^3 -3x^2 + 9x -9

Q4.f(x)=x+4 and g(x)=x+6

(f/g)(x) = f(x) ÷ g(x)

= x+4 / x+6

the domain restriction is x>-6

x<-6

x doesn't equal (-6)

7 0

3 years ago

HELP?

poizon [28]

My "work" is to enter the equations into a graphing calculator and let it show me the answers--just as your "work" is to copy the question to Brainly.

The solutions are
(x, y) = (-0.5, 7.5)
(x, y) = (3, 25)

_____
When you equate the expressions for y, you get the quadratic
2x² -5x -3 = 0
(2x +1)(x -3) = 0
x = -1/2 or 3
y = 5(x +2) = 15/2 or 25

4 0

3 years ago

Which equation describes the same line as y – 6 = –4(x + 1)?