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katovenus [111]
2 years ago
5

Paul developed a roll of film containing 36 pictures. If he made 2 prints each of half of the pictures and 1 print of each of th

e rest, how many prints did he make in all?
Mathematics
1 answer:
Monica [59]2 years ago
8 0

Answer:

54

Step-by-step explanation:

Total pictures in the roll = 36

half of the roll = 1/2 of the total pictures in the roll = 1/2 * 36 = 18

Given Paul made 2 prints of each of the half pictures

no. of prints made of half of the pictures = 18*2 = 36

No. of prints left after the half of the roll is printed for two prints= 36-18 = 18

No. of prints made of rest of the pictures 1

Therefore , no of prints made of rest half of pictures = 18*1 = 18

Total prints made = 36 + 18 = 54.

Thus, Paul made 54 prints in all.

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If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe
Natalka [10]

Answer:

They encircle the planet 3.76\times 10^{11} times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is 6.022\times 10^{23}

Thus, the number of dust particles is:

2.5\times 6.022\times 10^{23}=15.055\times 10^{23}

Diameter of a dust particles is 10μm and the circumference of earth is 40,076 km.

Convert the measurement in meters.

Diameter: 10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m

If we line up the particles the distance they could cover is:

15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}

Circumference in meters:

40,076km\times \frac{1000m}{1km}=40,076,000 m

Therefore,

\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}

Hence, they encircle the planet 3.76\times 10^{11} times.

8 0
2 years ago
AC ≅ AE; ∠ACD ≅ ∠AED
pshichka [43]

Answers:

A) △ACF ≅ △AEB because of ASA.

D) ∠CFA ≅ ∠EBA

E) FC ≅ BE


Solution:

AC ≅ AE; ∠ACD ≅ ∠AED Given

The angle ∠CAF ≅ ∠EAB, because is the same angle in Vertex A

Then △ACF ≅ △AEB because of ASA (Angle Side Angle): They have a congruent side (AC ≅ AE) and the two adjacent angles to this side are congruent too (∠ACD ≅ ∠AED and ∠CAF ≅ ∠EAB), then  option A)  is true: △ACF ≅ △AEB because of ASA.

If the two triangles are congruent, the ∠CFA ≅ ∠EBA; and FC ≅ BE, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), then Options D) ∠CFA ≅ ∠EBA and E) FC ≅ BE are true

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2 years ago
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13 times 4 is 52
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