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2 years ago

17.If -4 is zero of the polynomial p(x) = x2

Mathematics

1 answer:

Maksim231197 [3]2 years ago

3 0

Answer:

d) -1

hope it help

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-6x-11x=68 HEEEEEEEEEEELP

zheka24 [161]

Answer:

The correct answer would be 4. Explanation below

Step-by-step explanation:

-6x-11x = 68

-6x-11x = -17x

-17x = 68

68/-17 = 4

Hope this helps!

~Starr

7 0

2 years ago

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Which descriptions can describe more than one triangle? check all that apply

babunello [35]

Answer:

Angle measurements of 35°, 35°, and 110°

Angle measurements of 40°, 60°, and 80°

Step-by-step explanation:

If you specify only the three angles, you can have an infinite number of similar triangles of different sizes.

A is wrong. There can be only one triangle in which all three side lengths are specified.

C is wrong. The angles do not add up to 180°.

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2 years ago

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Samantha Smith purchased a $50,000 policy for $1,050 per year. After ten years, what will be the ratio (percentage) of

frosja888 [35]

Answer:

After 10 years the ratio will be 10500: 500000

4 0

2 years ago

Emily is a diver. Today she descended to a

kotegsom [21]

She was at 40 feet below, she swam up getting so she was then at 40-10 = 30 feet below sea level.

She the swan 4 feet down so she ended up at 30 + 4 = 34 feet below sea level.

The answer is 34 feet below sea level (-34)

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2 years ago

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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.