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vagabundo [1.1K]

2 years ago

12

What is the missing coordinate of the square with corners at (4, 0), (6, 0) and (6, 2)?

2 answers:

Vedmedyk [2.9K]2 years ago

7 0

(4,2) the first coordinate is 4, the second 2

Archy [21]2 years ago

6 0

I think the answer is (4,2)

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–2.5p – 20 = 9p + 37.5

Elenna [48]

Answer:

p = - 5

Step-by-step explanation:

–2.5p – 20 = 9p + 37.5

combine like terms:

- 2.5p - 9p = 37.5 + 20

simplify:

- 11.5p = 57.5

p = 57.5 / -11.5

p = - 5

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Brrunno [24]

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makvit [3.9K]

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A particle in the first quadrant is moving along a path described by the equation LaTeX: x^2+xy+2y^2=16x 2 + x y + 2 y 2 = 16 su

ElenaW [278]

Answer:

$\frac{50}{3}$ cm/sec.

Step-by-step explanation:

We have been given that a particle in the first quadrant is moving along a path described by the equation $x^2+xy+2y^2=16$ such that at the moment its x-coordinate is 2, its y-coordinate is decreasing at a rate of 10 cm/sec. We are asked to find the rate at which x-coordinate is changing at that time.

First of all, we will find the y value, when $x =2$ by substituting $x =2$ in our given equation.

$2^2+2y+2y^2=16$

$4-16+2y+2y^2=16-16$

$2y^2+2y-12=0$

$y^2+y-6=0$

$y^2+3y-2y-6=0$

$(y+3)(y-2)=0$

$(y+3)=0,(y-2)=0$

$y=-3,y=2$

Since the particle is moving in the 1st quadrant, so the value of y will be positive that is $y=2$ .

Now, we will find the derivative of our given equation.

$2x\cdot x'+x'y+xy'+4y\cdot y'=0$

We have been given that $y=2$ , $x =2$ and $y'=-10$ .

$2(2)\cdot x'+(2)x'+2(-10)+4(2)\cdot (-10)=0$

$4\cdot x'+2x'-20-80=0$

$6x'-100=0$

$6x'-100+100=0+100$

$6x'=100$

$\frac{6x'}{6}=\frac{100}{6}$

$x'=\frac{50}{3}$

Therefore, the x-coordinate is increasing at a rate of $\frac{50}{3}$ cm/sec.

7 0

3 years ago