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I am Lyosha [343]
3 years ago
15

Reflecting over the x-axis in words and formula

Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0

Answer:

Formula: (x, -y), when reflecting over the x-axis you keep the x-value the same, but change the sign of the y-value. For example: You have the original coordinates (3, 5), and if you reflect it over the x-axis it’ll be (3, -5).

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Mary and tom were buying a present for their mother on mothers day. mary spent $12 and tom spent $28 if they wanted to share the
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16 dollars is how much mary owes tom

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∫∫(x+y)dxdy ,d là miền giới hạn bởi x²+y²=1
igor_vitrenko [27]

It looks like you want to compute the double integral

\displaystyle \iint_D (x+y) \,\mathrm dx\,\mathrm dy

over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.

Convert to polar coordinates, in which <em>D</em> is given by the set

<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}

and

<em>x</em> = <em>r</em> cos(<em>θ</em>)

<em>y</em> = <em>r</em> sin(<em>θ</em>)

d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>

Then the integral is

\displaystyle \iint_D (x+y)\,\mathrm dx\,\mathrm dy = \iint_D r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \int_0^{2\pi} \int_0^1 r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \underbrace{\left( \int_0^{2\pi}(\cos(\theta)+\sin(\theta))\,\mathrm d\theta \right)}_{\int = 0} \left( \int_0^1 r^2\,\mathrm dr \right) = \boxed{0}

3 0
3 years ago
Help plz I’m confused
Ilia_Sergeevich [38]

Answer:

C.

Step-by-step explanation:

\frac{2}{3} is between 0 and 1

Only 1 point is between 0 and 1

C.

6 0
3 years ago
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Graph the line that passes through the points (-4, -1) and (-4, 2) and determine the equation of the line.​
My name is Ann [436]

Answer:

X = -4

Step-by-step explanation:

7 0
3 years ago
Tom put 18 gallons of mid grade gas in his truck and filled up his empty five gallon gas tank with regular gas for his lawn mowe
LuckyWell [14K]

Answer:

Mid grade gas per gallon = $2.67

Regular gas per gallon = $2.37

Step-by-step explanation:

Let, x= mid grade gas, y=regular gas.

So, for 18 gallons of mid grade gas and 5 gallon of regular gas at $59.91, it can be expressed as,

18x+5y=59.91  ------------------(equation 1)

for 14 gallon of mid grade gas and 1 gallon of regular gas at 39.75, it can be expressed as,

14x+y=39.75  -------------------(equation 2)

y=39.75-14x  -------------------(equation 3)

Now substituting the value of y from (equation 3) in (equation 1) we get,

18x+5y=59.91

18x+5(39.75-14x)=59.91

18x+198.75-70x=59.91

52x=198.75-59.91

52x=138.84

x=\frac{138.84}{52}

x=2.67  -------------------------(equation 4)

Now substituting value of x from (equation 4) in (equation 3) we get,

y=39.75-14x

y=39.75-(14\times2.67)

y=39.75-37.38

y=2.37

Therefore price per gallon of mid grade gas is x = $2.67, and price per gallon of regular gas y = $2.37.

4 0
3 years ago
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