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3 years ago

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Which of the following correctly uses absolute value to show the distance between -50 and 12? A: |-50 - 12| = |-62| = -62 units

B: |-50 + 12| = |-38| = 38 units C: |-50 + 12| = |-38| = -38 units D: |-50 - 12| = |-62| = 62 units

1 answer:

Zanzabum3 years ago

7 0

I’m sorry I’m answering this to ask a question

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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

Find the value of x for which ABCD must be a parallelogram.

kolbaska11 [484]

4x-1=x+26
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3 years ago

Read 2 more answers

The function y = f(x) is graphed below. What is the average rate of change of the

Serjik [45]

The average rate of change of the function in the given interval -3 ≤ x ≤-1 is 10.

<h3>What is the average rate of change?</h3>

The average Rate of Change of the function f(x) can be calculated as;

$f(x) = \dfrac{f(b) - f(a)}{b-a}$

Interval -3 ≤ x ≤-1

a = -3, b = -1

f(a) = f(-3) = -20

f(b) = f(-1) = 0

Step 2: Find Average

$f(x) = \dfrac{f(b) - f(a)}{b-a}\\\\\\f(x) = \dfrac{f(-1) - f(-3)}{-1-(-3)}\\\\\\f(x) = \dfrac{0 - (-20)}{2}\\\\f(x) = 10$

Therefore, the average rate of change of the function in the given interval -3 ≤ x ≤-1 is 10.

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Velocity time graph represent and exponential function

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