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2 years ago

What is 5/6 times 4 1/2 times 2 1/5

Mathematics

2 answers:

Levart [38]2 years ago

4 0

The answer to the problem is 8.25

RSB [31]2 years ago

3 0

5/6 x 9/2 x 11/5 = 495/ 60 = 8.25

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What is a whole number that can go in 30 and not in 15

zepelin [54]

Ten us a whole number that can go into 30 (30÷10=3) but not 15 (15÷10=1.5).

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2 years ago

Crop yield is the ratio of the number of bushels harvested to the number of acres used for the harvest. This year, the harvest a

yulyashka [42]

The total harvested area is 232 acres .

To determine the number of acres harvested, we simply have to divide the bushels of soybean by the crop yield.

Mathematically we can write it as,

<h3>What is the formula for the total land area?</h3>

The total land area = bushels of soybean / bushels per acre

Substituting the known values,

$Total land area = 9396 / 40.5 = 232$

Therefore, 232 acres were harvested.

To learn more about the area visit:

brainly.com/question/25292087

3 0

1 year ago

What is the square root of 12?<br>

andrew11 [14]

I beilieve it’s 11 I’m not 100% certain though

4 0

2 years ago

Read 2 more answers

The table shows the values of a function f(x).

QveST [7]

Answer:

The average rate of change is $=-4$

Step-by-step explanation:

The average rate of change from x=a to x=b of f(x) is given by;

$=\frac{f(b)-f(a)}{b-a}$

We want to find the average rate of change of the function represented in the table from x=-2 to x=2.

$=\frac{f(2)-f(-2)}{2--2}$

From the table f(-2)=25 and f(2)=9

The average rate of change from x=-2 to x=2 is

$=\frac{9-25}{2--2}$

$=\frac{-16}{4}$

$=-4$

3 0

2 years ago

Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v

k0ka [10]

I'll use subscript notation for brevity, i.e. $\frac{\partial f}{\partial x}=f_x$ .

By the chain rule,

$z_u=z_xx_u+z_yy_u$

$z_v=z_xx_v+z_yy_v$

$z_w=z_xx_w+z_yy_w$

We have

$z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}$

and

$\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}$

When $u=1,v=1,w=0$ , we have

$\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}$

and the partial derivatives take on values of

$\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}$

So we end up with

$\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}$

3 0

2 years ago