A cube root is representaive of reverse mutiplication. E.g Square root of 4 is 2 (2x2)
Using this, we can see that anytime 3 negatives are multiplied, the answer will always be negative.
<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>-</em><em>1</em><em>3</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>
<em>G</em><em>ood</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>
Answer:
1) 12
2) 106
3) 42
Step-by-step explanation:
Solve using PEMDAS.
Hope this helps. Pls give brainliest.
Answer: 3.66 cm
Step-by-step explanation: Given a rectangular casing BCDE with segment DE = 3 cm and segment BE = 3.5 cm.
The area A of a rectangle is length multiply by width.
Where length L = 3.5 cm and
width W = 3 cm
Area A = 3.5 × 3 = 10.5 cm^2
The pipe that will fit the fiber optic line is in cylindrical shape. Where area of a cylinder = πr^2.
But area A = 10.5. Substitute the values for the area of the cylinder
10.5 = πr^2
10.5 = 3.143 × r^2
Make r^2 the subject of formula
r^2 = 10.5/3.143
r = sqrt ( 3.34225 )
r = 1.828
Diameter = 2 × radius
Diameter = 2 × 1.829
Diameter = 3.656 cm
Therefore, the smallest diameter of pipe that will fit the fiber optic line is 3.66 cm approximately.
-2/9
3/9 * -2/3
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