Answer:
3.48cm³ (approximated)
Step-by-step explanation:
The maximum water that bowl can hold = The volume of the bowl
Radius is half of diameter.
<u>Given</u> -
<em>From</em><em> </em><em>the</em><em> </em><em>given</em><em> </em><em>data</em><em> </em><em>we</em><em> </em><em>can</em><em> </em><em>derive </em><em>that</em><em> </em><em>the</em><em> </em><em>radius</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>hemispheri</em><em>cal</em><em> </em><em>bowl</em><em> </em><em>is</em><em> </em><em>5</em><em>c</em><em>m</em><em> </em><em>(</em><em>1</em><em>0</em><em>/</em><em>2</em><em>)</em>
Putting the known values into the formula to get the value of the volume of bowl :
= 3.48888cm³
= 3.48cm³
It is an acute triangle because none of its angles are equal to (right) or above (obtuse) 90 degrees.
5/9 = .55555555555 yards traveled per hop
15/9 = 1.6666667 feet per hop
The image of the given scenario is attached below.
A right angled triangle is formed, with one angle equal to 27 degrees. The perpendicular side is 155 and we are to find the length of guy wire which makes the hypotenuse of the right angled triangle.
Using the formula of sine, we can write:
Rounding to nearest whole number, the length of the guy wire that must be attached is 341 feet.
Answer:
1,2, and 4
Step-by-step explanation:
I attached a picture with the graph
y= -4 is a straight horizontal line (red line) and
y≤10 will give a horizontal line passing trough y=10 (blue line)
the solution is all points under and on the line y=10
1, 2, and 4 are part of the solution set
(the blue area is the solution of the system of equations)
