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3 years ago

A canoe team leaves the dock at a bearing of 25° south

Mathematics

1 answer:

balandron [24]3 years ago

8 0

Answer:

ijnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

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The circle below is centered at (4, -1) and has a radius of 3. What is its

denis23 [38]

Answer:

Answers C and D are proably missing the y term. I'll assume they are:

C. (x-4)² + (y-1)² = 9

D. (x-4)² + (y-1)² = 3

A, is the equation that produces the drawn circle.

Step-by-step explanation:

See the attached image for explanation.

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2 years ago

Find the diffrence between temperature -30 degrees c and 70 degrees c?

andreyandreev [35.5K]

One is colder than the other.
Hope this helps :)

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3 years ago

Read 2 more answers

Rewrite 1/100,000,000 as a power of 10

yarga [219]

1/100,000,000 , note that 100,000,000 = 10⁸ , then:

1/100,000,000 = 1/10⁸
Remember that 1/aⁿ = a⁻ⁿ

Hence 1/100,000,000 = 1/10⁸ = 10⁻⁸

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3 years ago

Read 2 more answers

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

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3 years ago

F(x) = 2(x - 3)^2 - 2

Natali5045456 [20]

Hi. I was unsure of what exactly you wanted from this equation, so here's a quick analysis:

Domain: (-∞, ∞)

Range: (-2, ∞)

X-intercepts: (4, 0), (2, 0)

Y-intercept: (0, 16)

Axis of Symmetry: x = 3

Minimum value (vertex): (3, -2)

Standard form: y = 2x^2 - 12x + 16

4 0

3 years ago