Brainlest answer!!!!!!

I need a answer quick!!!! Find 1/10 of 1/100 of a meter

Tema [17]

1/10= .1 of a meter
1/10 of 1/100= .001 of a meter

3 0

2 years ago

Which equation has only one solution?

notka56 [123]

Absolute value cannot be less than 0
Solve Absolute Value.|x−5|=−1No solutions.

|−6−2x|=8
x=−7 or x=1

|5x+10|=10
x=0 or x=−4

|−6x+3|=0
 $x = \frac{1}{2}$

So your answer is D) |–6x + 3| = 0

6 0

2 years ago

A juice drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of drink on an average. Any overfill

ozzi

Answer:

d. The average is equal to 12 ounces.

Step-by-step explanation:

In this problem, the drink filling machine must be perfectly calibrated at 12 ounces since it needs to be shut down in cases of overfilling (mean > 12 ounces) and underfilling (mean < 12 ounces). Therefore, the correct approach would be to test if the mean is 12 ounces and the correct set of hypothesis would be:

$H_0:\mu=12\\H_a:\mu\neq 12$

The correct alternative is d. The average is equal to 12 ounces.

5 0

2 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