0.36x10=36.363636
0.36x1000=3636.363636
3636.36-36.36=3600
1000x-10x=990
3600/990=360/99
120/33=40/11 I think?
Answer:
From greatest to least
One and five-sixth rounded to 2
One and StartFraction 7 over 29 rounded to 1
0.16 rounded to 0
Step-by-step explanation:
1+7/29=29+7/29
=36/29
=1.24
Approximately
=1
1+5/6=6+5/6
=11/6
=1.83
Approximately
=2
0.16
Approximately =0
From greatest to least
One and five-sixth rounded to 2
One and StartFraction 7 over 29 rounded to 1
0.16 rounded to 0
Explanation:
There may be a more direct way to do this, but here's one way. We make no claim that the statements used here are on your menu of statements.
<u>Statement</u> . . . . <u>Reason</u>
2. ∆ADB, ∆ACB are isosceles . . . . definition of isosceles triangle
3. AD ≅ BD
and ∠CAE ≅ ∠CBE . . . . definition of isosceles triangle
4. ∠CAE = ∠CAD +∠DAE
and ∠CBE = ∠CBD +∠DBE . . . . angle addition postulate
5. ∠CAD +∠DAE ≅ ∠CBD +∠DBE . . . . substitution property of equality
6. ∠CAD +∠DAE ≅ ∠CBD +∠DAE . . . . substitution property of equality
7. ∠CAD ≅ ∠CBD . . . . subtraction property of equality
8. ∆CAD ≅ ∆CBD . . . . SAS congruence postulate
9. ∠ACD ≅ ∠BCD . . . . CPCTC
10. DC bisects ∠ACB . . . . definition of angle bisector
The original volume of the ballon, with radius r, is given by
V₀ = (4/3)πr³ in²
If the radius increases to (r+5) because of air input, the new volume is
V₁ = (4/3)π (r+5)³ in²
The difference in volume is the amount of air used to inflate the balloon.
The volume of air is
V = V₁ - V₀
= (4/3)π [r³ + 3r²(5) + 3r(5²) + 5³ - r³]
= (4/3)π (15r² + 75r + 125) in³
Answer: (4/3)π (15r² + 75r + 125) in³
Answer:
the function is an exponential funtion.
Step-by-step explanation:
learned it