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3 years ago

Simplify (8y)(-3y^2)

Mathematics

2 answers:

podryga [215]3 years ago

8 0

Answer:

$- 24 {y}^{3}$

Step-by-step explanation:

$(8y)( - 3 {y}^{2} )$

$= 8( - 3) {y}^{2} y$

$= - 24 {y}^{2 + 1}$

$= - 24 {y}^{3}$

Ainat [17]3 years ago

3 0

Good evening ,

______

Answer:

-24y⁴

___________________

Step-by-step explanation:

(8y)(-3y^2) = (8y)(-3y²) = -3×8×(y)×y³ = -24y⁴

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3 years ago

I need a answer fast please!

Paladinen [302]

Rewrite it without stem-and-leaf:

16,18,23,26,26,34,37,37,40,41,46

MEAN =∑(16,18,23,26,26,34,37,37,40,41,46)/11
MEAN =(16+18+23+25+26+34+37+37+40+41+46)/11 =343/11 = 31.18

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3 years ago

Barrett works at an ice cream shop. The function f(x) represents the amount of money in dollars Barrett earns per gallon of ice

Paha777 [63]

Answer:

$f(g(x))=6x^3} +4$

Step-by-step explanation:

Given:

$f(x)=2x^2+4\\\\g(x)=\sqrt{3x^3}$

Required:

f(g(x))=?

Solution:

let f(g(x))=f(X), where X=g(x)

so $f(X)=2X^2+4$

put X= $g(x)=\sqrt{3x^3}$ , we get

$f(g(x))=2(\sqrt{3x^3} )^2+4$

$f(g(x))=2(\sqrt{3x^3} )^2+4\\\\f(g(x))=2({3x^3} )+4\\\\f(g(x))=6x^3} +4$

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3 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

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Artyom0805 [142]

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