The area of a second soup can’s base is 5 pi inches squared. The can has a height of 10 inches. How would you use that informati
2 answers:
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Answer:
a). x = 11
b). m∠DMC = 39°
c). m∠MAD = 66°
d). m∠ADM = 36°
e). m∠ADC = 18°
Step-by-step explanation:
a). In the figure attached,
m∠AMC = 3x + 6
and m∠DMC = 6x - 49
Since "in-center" of a triangle is a points where the bisectors of internal angles meet.
Therefore, MC is the angle bisector of angle AMD.
and m∠AMC ≅ m∠DMC
3x + 6 = 8x - 49
8x - 3x = 49 + 6
5x = 55
x = 11
b). m∠DMC = 8x - 49
= (8 × 11) - 49
= 88 - 49
= 39°
c). m∠MAD = 2(m∠DAC)
= 2(30)°
= 60°
d). Since, m∠AMD + m∠ADM + m∠MAD = 180°
2(39)° + m∠ADM + 66° = 180°
78° + m∠ADM + 66° = 180°
m∠ADM = 180° - 144°
= 36°
e). m∠ADC =
=
= 18°
the upper bound for the length is .
<u>Step-by-step explanation:</u>
Lower and Upper Bounds
- The lower bound is the smallest value that will round up to the approximate value.
- The upper bound is the smallest value that will round up to the next approximate value.
Ex:- a mass of 70 kg, rounded to the nearest 10 kg, The upper bound is 75 kg, because 75 kg is the smallest mass that would round up to 80kg.
Here , A length is measured as 21cm correct to 2 significant figures. We need to find what is the upper bound for the length . let's find out:
As discussed above , upper bound for any number will be the smallest value in decimals which will round up to next integer value . So , for 21 :
⇒
21.5 cm on rounding off will give 22 cm . So , the upper bound for the length is .