Please answer this really fast.

Mathematics

2 answers:

solniwko [45]2 years ago

8 0

Answer:2x(3-1) +3(3x-1)

6x^2 -2x+9x-3

6x^2+7x-3

a=6 b=7 c=3

product (x)=6x7x-3=-126

dolphi86 [110]2 years ago

4 0

Answer:

abc = - 126

Step-by-step explanation:

(2x + 3)(3x - 1) ← expand product using FOIL

= 6x² + 7x - 3 ← in standard form

with a = 6, b = 7, c = - 3

Then

abc = 6 × 7 × (- 3) = - 126

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Graph y=-2/3x-3 i need help quick

timurjin [86]

Answer: just get gud at math lol

Step-by-step explanation: no

5 0

2 years ago

Consider the line integral Z C (sin x dx + cos y dy), where C consists of the top part of the circle x 2 + y 2 = 1 from (1, 0) t

pashok25 [27]

Direct computation:

Parameterize the top part of the circle $x^2+y^2=1$ by

$\vec r(t)=(x(t),y(t))=(\cos t,\sin t)$

with $0\le t\le\pi$ , and the line segment by

$\vec s(t)=(1-t)(-1,0)+t(2,-\pi)=(3t-1,-\pi t)$

with $0\le t\le1$ . Then

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)$

$=\displaystyle\int_0^\pi(-\sin t\sin(\cos t)+\cos t\cos(\sin t)\,\mathrm dt+\int_0^1(3\sin(3t-1)-\pi\cos(-\pi t))\,\mathrm dt$

$=0+(\cos1-\cos2)=\boxed{\cos1-\cos2}$

Using the fundamental theorem of calculus:

The integral can be written as

$\displaystyle\int_C(\sin x\,\mathrm dx+\cos y\,\mathrm dy)=\int_C\underbrace{(\sin x,\cos y)}_{\vec F}\cdot\underbrace{(\mathrm dx,\mathrm dy)}_{\vec r}$

If there happens to be a scalar function $f$ such that $\vec F=\nabla f$ , then $\vec F$ is conservative and the integral is path-independent, so we only need to worry about the value of $f$ at the path's endpoints.