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barxatty [35]
3 years ago
15

Please help meeeeeeeeeeeeeee

Mathematics
2 answers:
galina1969 [7]3 years ago
7 0
Together they scored:
(3x+2)+(300-2x)=347
x+302=347
x=45
Substituting x:
3*45+2=135+2=137
300-2*45=300-90=210
So Makayla scored a total of 210 points
lions [1.4K]3 years ago
4 0
She scored a total of 210 points
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Murljashka [212]
If Shelly is 3 years older than Michael and There 67. Michael is 64 and Shelly is 67. Or Michael is 67 and Shelly is 70.

Hope I helped. Have a wonderful day
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Dmitriy789 [7]

Answer:

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Step-by-step explanation:

7 0
3 years ago
The path of a golf ball can be modeled by the quadratic equation y=-0.001x^2+0.3x, where y represents the vertical distance and
Bingel [31]
Well if y represents the vertical distance of the ball then you want to solve the equation for when y is 0, since 0 would represent the ball being on the ground.  So substitute 0 in for y and solve the equation.

0 = - 0.001x^2 + 0.3x

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0 = x(-0.001x + 0.3)

Zero product property tells us that we can set both of those products equal to zero and solve

0 = x
0 = -0.001x + 0.3

our first answer is x = 0 is obvious bc thats when the ball starts before you hit it, means it traveled 0.

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8 0
3 years ago
37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the
Nastasia [14]

I'll only look at (37) here, since

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

In both parts, we're considering the line integral

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy

and I assume <em>C</em> has a positive orientation in both cases

(a) It looks like the region has the curves <em>y</em> = <em>x</em> and <em>y</em> = <em>x</em> ² as its boundary***, so that the interior of <em>C</em> is the set <em>D</em> given by

D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}

• Compute the line integral directly by splitting up <em>C</em> into two component curves,

<em>C₁ </em>: <em>x</em> = <em>t</em> and <em>y</em> = <em>t</em> ² with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}

(b) <em>C</em> is the boundary of the region

D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}

• Compute the line integral directly, splitting up <em>C</em> into 3 components,

<em>C₁</em> : <em>x</em> = <em>t</em> and <em>y</em> = 0 with 0 ≤ <em>t</em> ≤ 1

<em>C₂</em> : <em>x</em> = 1 - <em>t</em> and <em>y</em> = <em>t</em> with 0 ≤ <em>t</em> ≤ 1

<em>C₃</em> : <em>x</em> = 0 and <em>y</em> = 1 - <em>t</em> with 0 ≤ <em>t</em> ≤ 1

Then

\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}

• Using Green's theorem:

\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}

4 0
3 years ago
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