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3 years ago

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The area of the pacific ocean is approximatley 6.4x10^7 square miles. The area of the gulf of mexico is approximately 10^5 squar

e miles. How many times greater is the area of the pacific ocean than the area of the gulf of mexico

1 answer:

matrenka [14]3 years ago

8 0

Answer:

640 times greater

Step-by-step explanation:

Given the following

area of the pacific ocean = 6.4x10^7 square miles

area of the gulf of mexico = 10^5 square miles

Taking the ratio of the distances

pacific ocean/gulf of mexico = 6.4x10^7/10^5

pacific ocean/gulf of mexico = 6.4 * 10^2

pacific ocean/gulf of mexico =640

pacific ocean = 640gulf of mexico

Hence the area of the pacific ocean us 640 times greater than the area of the gulf of mexico

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32. Parallelogram ABCD has vertices A(0,0), B(2,4), and C(10,4). Find the coordinates of D.

goblinko [34]

Answer:

The fourth vertex is D(8, 0)

Step-by-step explanation:

Let A(0, 0), B(2, 4), and C(10, 4) be the three vertices of a parallelogram ABCD and let its fourth vertex be D(a, b).

Join AC and BD. Let AC and BD intersect at point O.

We have known that the diagonals of a parallelogram bisect each other.

So, O would be the midpoint of AC as well as that of BD.

$A\left(0,\:0\right)=\left(x_1,\:x_2\:\right)$

$C\left(10,\:4\right)=\left(x_2,\:x_2\:\right)$

The midpoint of AC is:

$\mathrm{Midpoint\:of\:}\left(x_1,\:y_1\right),\:\left(x_2,\:y_2\right):\quad \left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)$

$\left(x_1,\:y_1\right)=\left(0,\:0\right),\:\left(x_2,\:y_2\right)=\left(10,\:4\right)$

$=\left(\frac{10+0}{2},\:\frac{4+0}{2}\right)$

$=\left(5,\:2\right)$

The midpoint of BD is:

$=\left(\frac{a+2}{2},\:\frac{b+4}{2}\right)$

so

∵ $\frac{a+2}{2}=5$

$a+2=10$

$a=8$

∵ $\frac{b+4}{2}=2$

$b+4=4$

$b=0$

Hence, the fourth vertex is D(8, 0)

4 0

3 years ago

Find the line integral of f? around the perimeter of the rectangle with corners (3,0, (3,2, (?2,2, (?2,0, traversed in that orde

satela [25.4K]

Without knowing exactly what $f$ is, this is impossible to do. So let's assume $f(x,y)=1$ . Then the line integral over the given rectangle will correspond to the "signed" perimeter of the region.

You don't specify that the loop is complete, so in fact the integral will only give the "signed" length of three sides.

Parameterize the region by first partitioning the contour into three sub-contours:

$C_1:\mathbf r_1(t)=(3,0)(1-t)+(3,2)t=(3,2t)\implies\dfrac{\mathrm d\mathbf r_1}{\mathrm dt}=(0,2)$
$C_2:\mathbf r_2(t)=(3,2)(1-t)+(-2,2)t=(3-5t,2)\implies\dfrac{\mathrm d\mathbf r_2}{\mathrm dt}=(-5,0)$
$C_3:\mathbf r_3(t)=(-2,2)(1-t)+(-2,0)t=(-2,2-2t)\implies\dfrac{\mathrm d\mathbf r_3}{\mathrm dt}=(0,-2)$

where $0\le t\le1$ for each sub-contour. Then the line integral is given by

$\displaystyle\int_Cf\,\mathrm dS=\int_{C_i}f(\mathbf r_i(t))\cdot\frac{\mathrm d\mathbf r_i}{\mathrm dt}$

with $i\in\{1,2,3\}$ . You have

$\displaystyle\int_{C_1}f\,\mathrm dS=\int_0^1(1,1)\cdot(0,2)\,\mathrm dt=2$
$\displaystyle\int_{C_2}f\,\mathrm dS=\int_0^1(1,1)\cdot(5,0)\,\mathrm dt=5$
$\displaystyle\int_{C_1}f\,\mathrm dS=\int_0^1(1,1)\cdot(0,-2)\,\mathrm dt=-2$

Then the integral over the entire contour would be $2+5-2=5$ . Note that if the loop is complete, then the last leg of the contour would evaluate to -5, and so the total would end up as 0. This result would also follow from the fact that $f(x,y)$ is conservative, i.e. $f(x,y)=\nabla g(x,y)$ for some scalar field $g$ , and so the line integral is path independent. Its value would depend only on the endpoints of the contour, which in the case of a closed loop would simply be 0.

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3 years ago

A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value

Elden [556K]

Answer:

The launch angle should be adjusted to 30.63°

Step-by-step explanation:

The range of a projectile which is the horizontal distance covered by the projectile can be expressed as;

R =(v^2 sin2θ)/g

Where

R = range

v = initial speed

θ = launch angle

g = acceleration due to gravity

For the case above. When the projectile is launched at angle 13° above the horizontal.

θ1 = 13

R1 = (v^2 sin2θ1)/g

R1 = (v^2 sin26°)/g ....1

For the range to double

R2 = (v^2 sin2θ)/g .....2

R2 = 2R1

Substituting R2 and R1

(v^2 sin2θ)/g = 2 × (v^2 sin26°)/g

Divide both sides by v^2/g

sin2θ = 2sin26

2θ = sininverse(2sin26)

θ = sininverse(2sin26)/2

θ = 30.63°

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3 years ago

Could you awnser this please

aleksklad [387]

No it is not linear.

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3 years ago

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Answer:

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Step-by-step explanation:

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