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weqwewe [10]
2 years ago
10

There are 24 pictures on a roll of film. Adelle had 84 pictures from her trip home. She said that 12 other pictures were spoiled

. How many rolls of film did Adelle use?
Mathematics
1 answer:
Anna71 [15]2 years ago
7 0

Answer:

Total pictures = 84+12 = 96

roll of films = 96 / 24 = 4

she used 4 rolls

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The $2 are exchanged for 10 dimes and 4 quarters, which make 14 coins total.
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# of terms: 2

inConstant(s): might be wrong but i think its 5

Coefficient(s): might be y itself but i dont know

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I did what I could but this number is not factorable with rational numbers

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How do you write 5,392,029,004 in expanded form
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5,000,000,000 + 300,000,000 + 90,000,000 + 2,000,000 + 20,000 + 9,000 + 4
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3 years ago
Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

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3 years ago
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Savatey [412]
Hey there! :D

Since all the terms are being added together, and there is a common denominator with the fractions, we can just add them.

x+x= 2x  3/4+2/4= 5/4

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I hope this helps!
~kaikers
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