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3 years ago

13

What is the answer someone help meeee

1 answer:

Varvara68 [4.7K]3 years ago

6 0

Answer:

1) 8n-3

2) 3+(8-n)

3) 3(n-8)

4) 3(n+8)

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15-3(2+6 times -3) <br> What would this simplify to?

pickupchik [31]

-45 because of PEMDAS

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3 years ago

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i am a number beetween 20 and 30 who is divisible by 2 4 and 7. I am an even number as well who am I?

miskamm [114]

Answer:

28

Step-by-step explanation:

28/2=14

28/4=7

28/7=4

28 is an even number

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Thirty-five less than the square of a number is the same as 2 times the number. Find the number

Nookie1986 [14]

9514 1404 393

Answer:

-5 or 7

Step-by-step explanation:

Let x represent the number. The problem statement tells us ...

x^2 -35 = 2x

x^2 -2x -35 = 0 . . . . subtract 2x to put in standard form

(x -7)(x +5) = 0 . . . . . factor

x = 7 or x = -5 . . . . solutions are values of x that make the factors zero

The number could be either -5 or +7.

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2 years ago

21x21 I need real help with this.Really struggling ;)

Roman55 [17]

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9 months ago

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