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Tcecarenko [31]
3 years ago
5

What is the first step in simplifying the following expression: 2 + 3(4 + 5×2) − 8 + 3^2

Mathematics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

5 x 2 inside the parenthesis

Step-by-step explanation:

We need an order of operations to ensure we always arrive at the correct answer. It gives us a consistent way to work with numbers. We use the mnemonic device like PEMDAS to remember the correct order.

P-parenthesis

E-exponents

M-multiplication

D-division

A-add

S-subtract.

We apply them left to right doing inner operations before outer operations.

There fore our first step is in the parenthesis and multiplication because it is the inner most operations. 5 x 2 inside the parenthesis.

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Answer:

-6x+5y+8

Step-by-step explanation: subtract 9xfrom 3x

Subtract 2 from 10

-6x+5y+10-2

-6x+5y+8

6 0
3 years ago
A retired math teacher takes his niece shopping. He tells his niece that they can spend x dollars and x minus 49.5 equals 15. Wh
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X - 49.5 = 15
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3 years ago
Quick help with math please? 50 points.
Elden [556K]

y≤-2x +2

y< 1/3x +2

this is the answer to this

7 0
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Read 2 more answers
A bookshelf is 3 1/2 feet long.each book on the shelf is 5/8 inches wide. how many books will fit on the shelf
viktelen [127]
Number of books = 42  /    5/8       (Note 3 1/2 feet =  42 ins.)

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4 0
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

7 0
3 years ago
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