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Alexandra [31]
3 years ago
9

A plant grows at a constant rate of __ cm per day.

Mathematics
1 answer:
devlian [24]3 years ago
3 0

Plant grows at 3CM per day

equation: y= 3x

21 is y so 21= 3x. divide by 3 so it's been 7 days

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Find Q2 of the following data set.3, 9, 12, 4, 6, 40, 25, 14, 10
MatroZZZ [7]

So, the q2 would be 10

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7. Solve the equation.<br> x + 3 = 3x - 7<br> A-10<br> B - 12<br> 05<br> OD-2
VladimirAG [237]

Answer:

C. x = 5

Step-by-step explanation:

x + 3 = 3x - 7

Add -3x and -3 on both sides.

x - 3x = -7 -3

Add or subtract like terms.

-2x = -10

Multiply -1 on both sides.

2x = 10

Divide 2 into both sides.

2x/2 = 10/2

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7 0
3 years ago
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Find dy/dx for 4 - xy = y^3
storchak [24]

Answer:

\frac{dy}{dx}=-\frac{y}{3y^2+x}

Step-by-step explanation:

4-xy=y^3

dy/dx=?

\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}

Solving for dy/dx: Addind x dy/dx both sides of the equation:

-y-x\frac{dy}{dx}+x\frac{dy}{dx}=3y^2\frac{dy}{dx}+x\frac{dy}{dx} \\ -y=3y^2\frac{dy}{dx}+x\frac{dy}{dx}

Common factor dy/dx on the right side of the equation:

-y=(3y^2+x)\frac{dy}{dx}

Dividing both sides of the equation by 3y^2+x:

\frac{-y}{3y^2+x}=\frac{(3y^2+x)}{3y^2+x}\frac{dy}{dx}\\ -\frac{y}{3y^2+x}=\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{y}{3y^2+x}

7 0
3 years ago
Point′is the image of Q(3,−4) under a reflection across the x-axis.
gtnhenbr [62]

Answer:

      Q ' (3,4)

Step-by-step explanation:

When we reflect across the x axis, the y coordinate becomes -y

Q (3,-4) becomes Q' (3,- -4)

                              Q ' (3,4)

8 0
3 years ago
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