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4 years ago

Use the zero product property to find the solutions to the equation x^2-15x-100=0

Mathematics

2 answers:

7nadin3 [17]4 years ago

6 0

Answer:

$x = 20$ or $x = -5$

Step-by-step explanation:

To solve the quadratic equation we must factor the expression

Look for 2 numbers that when you multiply them, obtain the result -100 and when you add them, obtain the result -15.

You can verify that these numbers are -20 and 5

$-20 +5 = -15\\\\-20 * 5 = -100$

Then the polynomial factors are

$(x-20) (x + 5) = 0$

The zero product property says that if two terms $a * b = 0$ then

$a = 0$ or $b = 0$

$(x-20) = 0$ or $(x + 5) = 0$

$x = 20$ or $x = -5$

GarryVolchara [31]4 years ago

5 0

Answer:

x = -5 or x = 20

Step-by-step explanation:

There fore the answer is C

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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.