A picture frame can hold a photograph that is 16 inches long and 12 inches wide.

What was used to make the bread of the poor sumerians

AfilCa [17]

Bread made with barley, oats, or millet. The barley bread was sometimes used as a punishment.

8 0

3 years ago

A scientist is studying the growth of a particular species of plant. He writes the following equation to show the height of the

iogann1982 [59]

Answer:

The average of rate was of 0.4314 cm/day, which means that on average, during this interval, the plant grew by 0.4314 cm each day.

Step-by-step explanation:

Average rate of change:

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

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

In this question:

$f(x) = 12(1.03)^n$

From n = 3 to n = 10

$a = 3, b = 10$

$f(b) = f(10) = 12(1.03)^{10} = 16.13$

$f(a) = f(3) = 12(1.03)^{3} = 13.11$

Average rate of change:

$A = \frac{f(b)-f(a)}{b-a} = \frac{16.13-13.11}{10-3} = 0.4314$

The average of rate was of 0.4314 cm/day, which means that on average, during this interval, the plant grew by 0.4314 cm each day.

4 0

3 years ago

Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit. (Hint: Assume that

Vadim26 [7]

So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
that simply means, 3 circles with a radius of 3, overlapping the triangle

now, the area in the middle, the shaded one, will be, the whole area of the triangle MINUS those 3 circle sectors

hmmmm each sector has 60°, that means, all three of them will then be 60+60+60 or 180°, so the area of those three sectors, can be combined into a 180° sector, well, hell, 180° is really half a circle

so.... the area of those three sectors of 60° each, all three combined, is the same area of half a circle with a radius of 3

so $\bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
\textit{area of half a circle}\\\\
A=\cfrac{\pi r^2}{2}\\\\
-----------------------------\\\\$

$\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
\end{array}\impliedby \textit{area of shaded area}\\\\
-----------------------------\\\\
\boxed{\cfrac{36\sqrt{3}}{4}-\cfrac{9\pi }{2}}$

you can, add the fractions if you want, or leave them like that, or get their difference by using their decimal format