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3 years ago

This is 8th grade math 5y+2×5h=4h Solve for h

Mathematics

2 answers:

Zigmanuir [339]3 years ago

8 0

Answer:

h= −5/6y (negative 5 y over 6)

Step-by-step explanation:

5y+(2)(5)h=4h

Step 1: Add -4h to both sides.

10h+5y+−4h=4h+−4h

6h+5y=0

Step 2: Add -5y to both sides.

6h+5y+−5y=0+−5y

6h=−5y

Step 3: Divide both sides by 6.

6h/6=−5y/6

h=−5/6y

professor190 [17]3 years ago

3 0

Answer:

h = - 5y/ 6

Step-by-step explanation:

5y + 2 * 5h = 4h

Using BODMAS

Perform multiplication first

5y + 10h = 4h

Collect like terms

5y = 4h - 10h

5y = - 6h

Divide both sides by - 6

5y / - 6 = - 6h / - 6

h = - 5y/ 6

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3 years ago

Help me I have no idea how to do 18

sleet_krkn [62]

First, subtract px2 from both sides.

Now you have:

x3 - px2 = (1 - p) x1

Next, divide both sides by (1 - p)

So now you have

x3 - px2/(1 - p) = x1

...as your final answer

*You can decide if you want to leave the parenthesis in your final answer, I left them there so it could be visible where I put them. :)

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3 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2